If one has a sequence ${x_1, x_2, x_3, \ldots \in {\bf R}}$ of real numbers ${x_n}$, it is unambiguous what it means for that sequence to converge to a limit ${x \in {\bf R}}$: it means that for every ${\epsilon > 0}$, there exists an ${N}$ such that ${|x_n-x| \leq \epsilon}$ for all ${n > N}$. Similarly for a sequence ${z_1, z_2, z_3, \ldots \in {\bf C}}$ of complex numbers ${z_n}$ converging to a limit ${z \in {\bf C}}$.

More generally, if one has a sequence ${v_1, v_2, v_3, \ldots}$ of ${d}$-dimensional vectors ${v_n}$ in a real vector space ${{\bf R}^d}$ or complex vector space ${{\bf C}^d}$, it is also unambiguous what it means for that sequence to converge to a limit ${v \in {\bf R}^d}$ or ${v \in {\bf C}^d}$; it means that for every ${\epsilon > 0}$, there exists an ${N}$ such that ${\|v_n-v\| \leq \epsilon}$ for all ${n \geq N}$. Here, the norm ${\|v\|}$ of a vector ${v = (v^{(1)},\ldots,v^{(d)})}$ can be chosen to be the Euclidean norm ${\|v\|_2 := (\sum_{j=1}^d (v^{(j)})^2)^{1/2}}$, the supremum norm ${\|v\|_\infty := \sup_{1 \leq j \leq d} |v^{(j)}|}$, or any other number of norms, but for the purposes of convergence, these norms are all equivalent; a sequence of vectors converges in the Euclidean norm if and only if it converges in the supremum norm, and similarly for any other two norms on the finite-dimensional space ${{\bf R}^d}$ or ${{\bf C}^d}$.

If however one has a sequence ${f_1, f_2, f_3, \ldots}$ of functions ${f_n: X \rightarrow {\bf R}}$ or ${f_n: X \rightarrow {\bf C}}$ on a common domain ${X}$, and a putative limit ${f: X \rightarrow {\bf R}}$ or ${f: X \rightarrow {\bf C}}$, there can now be many different ways in which the sequence ${f_n}$ may or may not converge to the limit ${f}$. (One could also consider convergence of functions ${f_n: X_n \rightarrow {\bf C}}$ on different domains ${X_n}$, but we will not discuss this issue at all here.) This is contrast with the situation with scalars ${x_n}$ or ${z_n}$ (which corresponds to the case when ${X}$ is a single point) or vectors ${v_n}$ (which corresponds to the case when ${X}$ is a finite set such as ${\{1,\ldots,d\}}$). Once ${X}$ becomes infinite, the functions ${f_n}$ acquire an infinite number of degrees of freedom, and this allows them to approach ${f}$ in any number of inequivalent ways.

What different types of convergence are there? As an undergraduate, one learns of the following two basic modes of convergence:

1. We say that ${f_n}$ converges to ${f}$ pointwise if, for every ${x \in X}$, ${f_n(x)}$ converges to ${f(x)}$. In other words, for every ${\epsilon > 0}$ and ${x \in X}$, there exists ${N}$ (that depends on both ${\epsilon}$ and ${x}$) such that ${|f_n(x)-f(x)| \leq \epsilon}$ whenever ${n \geq N}$.
2. We say that ${f_n}$ converges to ${f}$ uniformly if, for every ${\epsilon > 0}$, there exists ${N}$ such that for every ${n \geq N}$, ${|f_n(x) - f(x)| \leq \epsilon}$ for every ${x \in X}$. The difference between uniform convergence and pointwise convergence is that with the former, the time ${N}$ at which ${f_n(x)}$ must be permanently ${\epsilon}$-close to ${f(x)}$ is not permitted to depend on ${x}$, but must instead be chosen uniformly in ${x}$.

Uniform convergence implies pointwise convergence, but not conversely. A typical example: the functions ${f_n: {\bf R} \rightarrow {\bf R}}$ defined by ${f_n(x) := x/n}$ converge pointwise to the zero function ${f(x) := 0}$, but not uniformly.

However, pointwise and uniform convergence are only two of dozens of many other modes of convergence that are of importance in analysis. We will not attempt to exhaustively enumerate these modes here (but see this Wikipedia page, and see also these 245B notes on strong and weak convergence). We will, however, discuss some of the modes of convergence that arise from measure theory, when the domain ${X}$ is equipped with the structure of a measure space ${(X, {\mathcal B}, \mu)}$, and the functions ${f_n}$ (and their limit ${f}$) are measurable with respect to this space. In this context, we have some additional modes of convergence:

1. We say that ${f_n}$ converges to ${f}$ pointwise almost everywhere if, for (${\mu}$-)almost everywhere ${x \in X}$, ${f_n(x)}$ converges to ${f(x)}$.
2. We say that ${f_n}$ converges to ${f}$ uniformly almost everywhere, essentially uniformly, or in ${L^\infty}$ norm if, for every ${\epsilon > 0}$, there exists ${N}$ such that for every ${n \geq N}$, ${|f_n(x) - f(x)| \leq \epsilon}$ for ${\mu}$-almost every ${x \in X}$.
3. We say that ${f_n}$ converges to ${f}$ almost uniformly if, for every ${\epsilon > 0}$, there exists an exceptional set ${E \in {\mathcal B}}$ of measure ${\mu(E) \leq \epsilon}$ such that ${f_n}$ converges uniformly to ${f}$ on the complement of ${E}$.
4. We say that ${f_n}$ converges to ${f}$ in ${L^1}$ norm if the quantity ${\|f_n-f\|_{L^1(\mu)} = \int_X |f_n(x)-f(x)|\ d\mu}$ converges to ${0}$ as ${n \rightarrow \infty}$.
5. We say that ${f_n}$ converges to ${f}$ in measure if, for every ${\epsilon > 0}$, the measures ${\mu( \{ x \in X: |f_n(x) - f(x)| \geq \epsilon \} )}$ converge to zero as ${n \rightarrow \infty}$.

Observe that each of these five modes of convergence is unaffected if one modifies ${f_n}$ or ${f}$ on a set of measure zero. In contrast, the pointwise and uniform modes of convergence can be affected if one modifies ${f_n}$ or ${f}$ even on a single point.

Remark 1 In the context of probability theory, in which ${f_n}$ and ${f}$ are interpreted as random variables, convergence in ${L^1}$ norm is often referred to as convergence in mean, pointwise convergence almost everywhere is often referred to as almost sure convergence, and convergence in measure is often referred to as convergence in probability.

Exercise 1 (Linearity of convergence) Let ${(X, {\mathcal B}, \mu)}$ be a measure space, let ${f_n, g_n: X \rightarrow {\bf C}}$ be sequences of measurable functions, and let ${f, g: X \rightarrow {\bf C}}$ be measurable functions.

1. Show that ${f_n}$ converges to ${f}$ along one of the above seven modes of convergence if and only if ${|f_n-f|}$ converges to ${0}$ along the same mode.
2. If ${f_n}$ converges to ${f}$ along one of the above seven modes of convergence, and ${g_n}$ converges to ${g}$ along the same mode, show that ${f_n+g_n}$ converges to ${f+g}$ along the same mode, and that ${cf_n}$ converges to ${cf}$ along the same mode for any ${c \in {\bf C}}$.
3. (Squeeze test) If ${f_n}$ converges to ${0}$ along one of the above seven modes, and ${|g_n| \leq f_n}$ pointwise for each ${n}$, show that ${g_n}$ converges to ${0}$ along the same mode.

We have some easy implications between modes:

Exercise 2 (Easy implications) Let ${(X, {\mathcal B}, \mu)}$ be a measure space, and let ${f_n: X \rightarrow {\bf C}}$ and ${f: X \rightarrow {\bf C}}$ be measurable functions.

1. If ${f_n}$ converges to ${f}$ uniformly, then ${f_n}$ converges to ${f}$ pointwise.
2. If ${f_n}$ converges to ${f}$ uniformly, then ${f_n}$ converges to ${f}$ in ${L^\infty}$ norm. Conversely, if ${f_n}$ converges to ${f}$ in ${L^\infty}$ norm, then ${f_n}$ converges to ${f}$ uniformly outside of a null set (i.e. there exists a null set ${E}$ such that the restriction ${f_n\downharpoonright_{X \backslash E}}$ of ${f_n}$ to the complement of ${E}$ converges to the restriction ${f\downharpoonright_{X \backslash E}}$ of ${f}$).
3. If ${f_n}$ converges to ${f}$ in ${L^\infty}$ norm, then ${f_n}$ converges to ${f}$ almost uniformly.
4. If ${f_n}$ converges to ${f}$ almost uniformly, then ${f_n}$ converges to ${f}$ pointwise almost everywhere.
5. If ${f_n}$ converges to ${f}$ pointwise, then ${f_n}$ converges to ${f}$ pointwise almost everywhere.
6. If ${f_n}$ converges to ${f}$ in ${L^1}$ norm, then ${f_n}$ converges to ${f}$ in measure.
7. If ${f_n}$ converges to ${f}$ almost uniformly, then ${f_n}$ converges to ${f}$ in measure.

The reader is encouraged to draw a diagram that summarises the logical implications between the seven modes of convergence that the above exercise describes.

We give four key examples that distinguish between these modes, in the case when ${X}$ is the real line ${{\bf R}}$ with Lebesgue measure. The first three of these examples already were introduced in the previous set of notes.

Example 1 (Escape to horizontal infinity) Let ${f_n := 1_{[n,n+1]}}$. Then ${f_n}$ converges to zero pointwise (and thus, pointwise almost everywhere), but not uniformly, in ${L^\infty}$ norm, almost uniformly, in ${L^1}$ norm, or in measure.

Example 2 (Escape to width infinity) Let ${f_n := \frac{1}{n} 1_{[0,n]}}$. Then ${f_n}$ converges to zero uniformly (and thus, pointwise, pointwise almost everywhere, in ${L^\infty}$ norm, almost uniformly, and in measure), but not in ${L^1}$ norm.

Example 3 (Escape to vertical infinity) Let ${f_n := n 1_{[\frac{1}{n}, \frac{2}{n}]}}$. Then ${f_n}$ converges to zero pointwise (and thus, pointwise almost everywhere) and almost uniformly (and hence in measure), but not uniformly, in ${L^\infty}$ norm, or in ${L^1}$ norm.

Example 4 (Typewriter sequence) Let ${f_n}$ be defined by the formula

$\displaystyle f_n := 1_{[\frac{n-2^k}{2^k}, \frac{n-2^k+1}{2^k}]}$

whenever ${k \geq 0}$ and ${2^k \leq n < 2^{k+1}}$. This is a sequence of indicator functions of intervals of decreasing length, marching across the unit interval ${[0,1]}$ over and over again. Then ${f_n}$ converges to zero in measure and in ${L^1}$ norm, but not pointwise almost everywhere (and hence also not pointwise, not almost uniformly, nor in ${L^\infty}$ norm, nor uniformly).

Remark 2 The ${L^\infty}$ norm ${\|f\|_{L^\infty(\mu)}}$ of a measurable function ${f: X \rightarrow {\bf C}}$ is defined to the infimum of all the quantities ${M \in [0,+\infty]}$ that are essential upper bounds for ${f}$ in the sense that ${|f(x)| \leq M}$ for almost every ${x}$. Then ${f_n}$ converges to ${f}$ in ${L^\infty}$ norm if and only if ${\|f_n-f\|_{L^\infty(\mu)} \rightarrow 0}$ as ${n \rightarrow \infty}$. The ${L^\infty}$ and ${L^1}$ norms are part of the larger family of ${L^p}$ norms, which we will study in more detail in 245B.

One particular advantage of ${L^1}$ convergence is that, in the case when the ${f_n}$ are absolutely integrable, it implies convergence of the integrals,

$\displaystyle \int_X f_n\ d\mu \rightarrow \int_X f\ d\mu,$

as one sees from the triangle inequality. Unfortunately, none of the other modes of convergence automatically imply this convergence of the integral, as the above examples show.

The purpose of these notes is to compare these modes of convergence with each other. Unfortunately, the relationship between these modes is not particularly simple; unlike the situation with pointwise and uniform convergence, one cannot simply rank these modes in a linear order from strongest to weakest. This is ultimately because the different modes react in different ways to the three “escape to infinity” scenarios described above, as well as to the “typewriter” behaviour when a single set is “overwritten” many times. On the other hand, if one imposes some additional assumptions to shut down one or more of these escape to infinity scenarios, such as a finite measure hypothesis ${\mu(X) < \infty}$ or a uniform integrability hypothesis, then one can obtain some additional implications between the different modes.

— 1. Uniqueness —

Throughout these notes, ${(X,{\mathcal B},\mu)}$ denotes a measure space. We abbreviate “${\mu}$-almost everywhere” as “almost everywhere” throughout.

Even though the modes of convergence all differ from each other, they are all compatible in the sense that they never disagree about which function ${f}$ a sequence of functions ${f_n}$ converges to, outside of a set of measure zero. More precisely:

Proposition 1 Let ${f_n: X \rightarrow {\bf C}}$ be a sequence of measurable functions, and let ${f, g: X \rightarrow {\bf C}}$ be two additional measurable functions. Suppose that ${f_n}$ converges to ${f}$ along one of the seven modes of convergence defined above, and ${f_n}$ converges to ${g}$ along another of the seven modes of convergence (or perhaps the same mode of convergence as for ${f}$). Then ${f}$ and ${g}$ agree almost everywhere.

Note that the conclusion is the best one can hope for in the case of the last five modes of convergence, since as remarked earlier, these modes of convergence are unaffected if one modifies ${f}$ or ${g}$ on a set of measure zero.

Proof: In view of Exercise 2, we may assume that ${f_n}$ converges to ${f}$ either pointwise almost everywhere, or in measure, and similarly that ${f_n}$ converges to ${g}$ either pointwise almost everywhere, or in measure.

Suppose first that ${f_n}$ converges to both ${f}$ and ${g}$ pointwise almost everywhere. Then by Exercise 1, ${0}$ converges to ${f-g}$ pointwise almost everywhere, which clearly implies that ${f-g}$ is zero almost everywhere, and the claim follows. A similar argument applies if ${f_n}$ converges to both ${f}$ and ${g}$ in measure.

By symmetry, the only remaining case that needs to be considered is when ${f_n}$ converges to ${f}$ pointwise almost everywhere, and ${f_n}$ converges to ${g}$ in measure. We need to show that ${f=g}$ almost everywhere. It suffices to show that for every ${\epsilon>0}$, that ${|f(x)-g(x)| \leq \epsilon}$ for almost every ${x}$, as the claim then follows by setting ${\epsilon = 1/m}$ for ${m=1,2,3,\ldots}$ and using the fact that the countable union of null sets is again a null set.

Fix ${\epsilon>0}$, and let ${A := \{ x \in X: |f(x)-g(x)| > \epsilon \}}$. This is a measurable set; our task is to show that it has measure zero. Suppose for contradiction that ${\mu(A) > 0}$. We consider the sets

$\displaystyle A_N := \{ x \in A: |f_n(x)-f(x)| \leq \epsilon/2 \hbox{ for all } n \geq N \}.$

These are measurable sets that are increasing in ${N}$. As ${f_n}$ converges to ${f}$ almost everywhere, we see that almost every ${x \in A}$ belongs to at least one of the ${A_N}$, thus ${\bigcup_{N=1}^\infty A_N}$ is equal to ${A}$ outside of a null set. In particular,

$\displaystyle \mu( \bigcup_{N=1}^\infty A_N ) > 0.$

Applying monotone convergence for sets, we conclude that

$\displaystyle \mu(A_N) > 0$

for some finite ${N}$. But by the triangle inequality, we have ${|f_n(x)-g(x)| > \epsilon/2}$ for all ${x \in A_N}$ and all ${n \geq N}$. As a consequence, ${f_n}$ cannot converge in measure to ${g}$, which gives the desired contradiction. $\Box$

— 2. The case of a step function —

One way to appreciate the distinctions between the above modes of convergence is to focus on the case when ${f=0}$, and when each of the ${f_n}$ is a step function, by which we mean a constant multiple ${f_n = A_n 1_{E_n}}$ of a measurable set ${E_n}$. For simplicity we will assume that the ${A_n > 0}$ are positive reals, and that the ${E_n}$ have a positive measure ${\mu(E_n) > 0}$. We also assume the ${A_n}$ exhibit one of two modes of behaviour: either the ${A_n}$ converge to zero, or else they are bounded away from zero (i.e. there exists ${c>0}$ such that ${A_n \geq c}$ for every ${n}$. It is easy to see that if a sequence ${A_n}$ does not converge to zero, then it has a subsequence that is bounded away from zero, so it does not cause too much loss of generality to restrict to one of these two cases.

Given such a sequence ${f_n = A_n 1_{E_n}}$ of step functions, we now ask, for each of the seven modes of convergence, what it means for this sequence to converge to zero along that mode. It turns out that the answer to question is controlled more or less completely by the following three quantities:

• The height ${A_n}$ of the ${n^{th}}$ function ${f_n}$;
• The width ${\mu(E_n)}$ of the ${n^{th}}$ function ${f_n}$; and
• The ${N^{th}}$ tail support ${E^*_N := \bigcup_{n \geq N} E_n}$ of the sequence ${f_1, f_2, f_3, \ldots}$.

Indeed, we have:

Exercise 3 (Convergence for step functions) Let the notation and assumptions be as above. Establish the following claims:

1. ${f_n}$ converges uniformly to zero if and only if ${A_n \rightarrow 0}$ as ${n \rightarrow \infty}$.
2. ${f_n}$ converges in ${L^\infty}$ norm to zero if and only if ${A_n \rightarrow 0}$ as ${n \rightarrow \infty}$.
3. ${f_n}$ converges almost uniformly to zero if and only if ${A_n \rightarrow 0}$ as ${n \rightarrow \infty}$, or ${\mu(E^*_N) \rightarrow 0}$ as ${N \rightarrow \infty}$.
4. ${f_n}$ converges pointwise to zero if and only if ${A_n \rightarrow 0}$ as ${n \rightarrow \infty}$, or ${\bigcap_{N=1}^\infty E^*_N = \emptyset}$.
5. ${f_n}$ converges pointwise almost everywhere to zero if and only if ${A_n \rightarrow 0}$ as ${n \rightarrow \infty}$, or ${\bigcap_{N=1}^\infty E^*_N}$ is a null set.
6. ${f_n}$ converges in measure to zero if and only if ${A_n \rightarrow 0}$ as ${n \rightarrow \infty}$, or ${\mu(E_n) \rightarrow 0}$ as ${n \rightarrow \infty}$.
7. ${f_n}$ converges in ${L^1}$ norm if and only if ${A_n \mu(E_n) \rightarrow 0}$ as ${n \rightarrow \infty}$.

To put it more informally: when the height goes to zero, then one has convergence to zero in all modes except possibly for ${L^1}$ convergence, which requires that the product of the height and the width goes to zero. If instead the height is bounded away from zero and the width is positive, then we never have uniform or ${L^\infty}$ convergence, but we have convergence in measure if the width goes to zero, we have almost uniform convergence if the tail support (which has larger measure than the width) has measure that goes to zero, we have pointwise almost everywhere convergence if the tail support shrinks to a null set, and pointwise convergence if the tail support shrinks to the empty set.

It is instructive to compare this exercise with Exercise 2, or with the four examples given in the introduction. In particular:

• In the escape to horizontal infinity scenario, the height and width do not shrink to zero, but the tail set shrinks to the empty set (while remaining of infinite measure throughout).
• In the escape to width infinity scenario, the height goes to zero, but the width (and tail support) go to infinity, causing the ${L^1}$ norm to stay bounded away from zero.
• In the escape to vertical infinity, the height goes to infinity, but the width (and tail support) go to zero (or the empty set), causing the ${L^1}$ norm to stay bounded away from zero.
• In the typewriter example, the width goes to zero, but the height and the tail support stay fixed (and thus bounded away from zero).

Remark 3 The monotone convergence theorem can also be specialised to this case. Observe that the ${f_n = A_n 1_{E_n}}$ are monotone increasing if and only if ${A_n \leq A_{n+1}}$ and ${E_n \subset E_{n+1}}$ for each ${n}$. In such cases, observe that the ${f_n}$ converge pointwise to ${f := A 1_E}$, where ${A := \lim_{n \rightarrow \infty} A_n}$ and ${E := \bigcup_{n=1}^\infty E_n}$. The monotone convergence theorem then asserts that ${A_n \mu(E_n) \rightarrow A \mu(E)}$ as ${n \rightarrow \infty}$, which is a consequence of the monotone convergence theorem ${\mu(E_n) \rightarrow \mu(E)}$ for sets.

— 3. Finite measure spaces —

The situation simplifies somewhat if the space ${X}$ has finite measure (and in particular, in the case when ${(X,{\mathcal B},\mu)}$ is a probability space). This shuts down two of the four examples (namely, escape to horizontal infinity or width infinity) and creates a few more equivalences. Indeed, from Egorov’s theorem (Exercise 30 from Notes 3), we now have

Theorem 2 (Egorov’s theorem, again) Let ${X}$ have finite measure, and let ${f_n: X \rightarrow {\bf C}}$ and ${f: X \rightarrow {\bf C}}$ be measurable functions. Then ${f_n}$ converges to ${f}$ pointwise almost everywhere if and only if ${f_n}$ converges to ${f}$ almost uniformly.

Note that when one specialises to step functions using Exercise 3, then Egorov’s theorem collapses to the downward monotone convergence property for sets (Exercise 23 of Notes 3).

Another nice feature of the finite measure case is that ${L^\infty}$ convergence implies ${L^1}$ convergence:

Exercise 4 Let ${X}$ have finite measure, and let ${f_n: X \rightarrow {\bf C}}$ and ${f: X \rightarrow {\bf C}}$ be measurable functions. Show that if ${f_n}$ converges to ${f}$ in ${L^\infty}$ norm, then ${f_n}$ also converges to ${f}$ in ${L^1}$ norm.

— 4. Fast convergence —

The typewriter example shows that ${L^1}$ convergence is not strong enough to force almost uniform or pointwise almost everywhere convergence. However, this can be rectified if one assumes that the ${L^1}$ convergence is sufficiently fast:

Exercise 5 (Fast ${L^1}$ convergence) Suppose that ${f_n, f: X \rightarrow {\bf C}}$ are measurable functions such that ${\sum_{n=1}^\infty \|f_n-f\|_{L^1(\mu)} < \infty}$; thus, not only do the quantities ${\|f_n-f\|_{L^1(\mu)}}$ go to zero (which would mean ${L^1}$ convergence), but they converge in an absolutely summable fashion.

1. Show that ${f_n}$ converges pointwise almost everywhere to ${f}$.
2. Show that ${f_n}$ converges almost uniformly to ${f}$.

(Hint: If you have trouble getting started, try working first in the special case in which ${f_n= A_n 1_{E_n}}$ are step functions and ${f=0}$ and use Exercise 3 in order to gain some intuition. The second part of the exercise implies the first, but the first is a little easier to prove and may thus serve as a useful warmup. The ${\epsilon/2^n}$ trick may come in handy for the second part.)

As a corollary, we see that ${L^1}$ convergence implies almost uniform or pointwise almost everywhere convergence if we are allowed to pass to a subsequence:

Corollary 3 Suppose that ${f_n: X \rightarrow {\bf C}}$ are a sequence of measurable functions that converge in ${L^1}$ norm to a limit ${f}$. Then there exists a subsequence ${f_{n_j}}$ that converges almost uniformly (and hence, pointwise almost everywhere) to ${f}$ (while remaining convergent in ${L^1}$ norm, of course).

Proof: Since ${\|f_n-f\|_{L^1(\mu)} \rightarrow 0}$ as ${n \rightarrow \infty}$, we can select ${n_1 < n_2 < n_3 < \ldots}$ such that ${\|f_{n_j} - f\|_{L^1(\mu)} \leq 2^{-j}}$ (say). This is enough for the previous exercise to apply. $\Box$

Actually, one can strengthen this corollary a bit by relaxing ${L^1}$ convergence to convergence in measure:

Exercise 6 Suppose that ${f_n: X \rightarrow {\bf C}}$ are a sequence of measurable functions that converge in measure to a limit ${f}$. Then there exists a subsequence ${f_{n_j}}$ that converges almost uniformly (and hence, pointwise almost everywhere) to ${f}$. (Hint: Choose the ${n_j}$ so that the sets ${\{ x \in X: |f_{n_j}(x) - f(x)| > 1/j \}}$ have a suitably small measure.)

It is instructive to see how this subsequence is extracted in the case of the typewriter sequence. In general, one can view the operation of passing to a subsequence as being able to eliminate “typewriter” situations in which the tail support is much larger than the width.

— 5. Domination and uniform integrability —

Now we turn to the reverse question, of whether almost uniform convergence, pointwise almost everywhere convergence, or convergence in measure can imply ${L^1}$ convergence. The escape to vertical and width infinity examples shows that without any further hypotheses, the answer to this question is no. However, one can do better if one places some domination hypotheses on the ${f_n}$ that shut down both of these escape routes.

We say that a sequence ${f_n: X \rightarrow {\bf C}}$ is dominated if there exists an absolutely integrable function ${g: X \rightarrow {\bf C}}$ such that ${|f_n(x)| \leq g(x)}$ for all ${n}$ and almost every ${x}$. For instance, if ${X}$ has finite measure and the ${f_n}$ are uniformly bounded, then they are dominated. Observe that the sequences in the vertical and width escape to infinity examples are not dominated (why?).

The dominated convergence theorem (Theorem 17 of Notes 3) then asserts that if ${f_n}$ converges to ${f}$ pointwise almost everywhere, then it necessarily converges to ${f}$ in ${L^1}$ norm (and hence also in measure). Here is a variant:

Exercise 7 Suppose that ${f_n: X \rightarrow {\bf C}}$ are a dominated sequence of measurable functions, and let ${f: X \rightarrow {\bf C}}$ be another measurable function. Show that ${f_n}$ converges in ${L^1}$ norm to ${f}$ if and only if ${f_n}$ converges in measure to ${f}$. (Hint: one way to establish the “if” direction is first show that every subsequence of the ${f_n}$ has a further subsequence that converges in ${L^1}$ to ${f}$, using Exercise 6 and the dominated convergence theorem. Alternatively, use monotone convergence to find a set ${E}$ of finite measure such that ${\int_{X \backslash E} g\ d\mu}$, and hence ${\int_{X \backslash E} f_n\ d\mu}$ and ${\int_{X \backslash E} f\ d\mu}$, are small.)

There is a more general notion than domination, known as uniform integrability, which serves as a substitute for domination in many (but not all) contexts.

Definition 4 (Uniform integrability) A sequence ${f_n: X \rightarrow {\bf C}}$ of absolutely integrable functions is said to be uniformly integrable if the following three statements hold:

• (Uniform bound on ${L^1}$ norm) One has ${\sup_n \|f_n\|_{L^1(\mu)} = \sup_n \int_X |f_n|\ d\mu < +\infty}$.
• (No escape to vertical infinity) One has ${\sup_n \int_{|f_n| \geq M} |f_n|\ d\mu \rightarrow 0}$ as ${M \rightarrow +\infty}$.
• (No escape to width infinity) One has ${\sup_n \int_{|f_n| \leq \delta} |f_n|\ d\mu \rightarrow 0}$ as ${\delta \rightarrow 0}$.

Remark 4 It is instructive to understand uniform integrability in the step function case ${f_n = A_n 1_{E_n}}$. The uniform bound on the ${L^1}$ norm then asserts that ${A_n \mu(E_n)}$ stays bounded. The lack of escape to vertical infinity means that along any subsequence for which ${A_n \rightarrow \infty}$, ${A_n \mu(E_n)}$ must go to zero. Similarly, the lack of escape to width infinity means that along any subsequence for which ${A_n \rightarrow 0}$, ${A_n \mu(E_n)}$ must go to zero.

Exercise 8

1. Show that if ${f}$ is an absolutely integrable function, then the constant sequence ${f_n=f}$ is uniformly integrable. (Hint: use the monotone convergence theorem.)
2. Show that every dominated sequence of measurable functions is uniformly integrable.
3. Give an example of a sequence that is uniformly integrable but not dominated.

In the case of a finite measure space, there is no escape to width infinity, and the criterion for uniform integrability simplifies to just that of excluding vertical infinity:

Exercise 9 Suppose that ${X}$ has finite measure, and let ${f_n: X \rightarrow {\bf C}}$ be a sequence of measurable functions. Show that ${f_n}$ is uniformly integrable if and only if ${\sup_n \int_{|f_n| \geq M} |f_n|\ d\mu \rightarrow 0}$ as ${M \rightarrow +\infty}$.

Exercise 10 (Uniform ${L^p}$ bound on finite measure implies uniform integrability) Suppose that ${X}$ have finite measure, let ${1 < p < \infty}$, an d suppose that ${f_n: X \rightarrow {\bf C}}$ is a sequence of measurable functions such that ${\sup_n \int_X |f_n|^p\ d\mu < \infty}$. Show that the sequence ${f_n}$ is uniformly integrable.

Exercise 11 Let ${f_n: X \rightarrow {\bf C}}$ be a uniformly integrable sequence of functions. Show that for every ${\epsilon > 0}$ there exists a ${\delta > 0}$ such that

$\displaystyle \int_E |f_n|\ d\mu \leq \epsilon$

whenever ${n \geq 1}$ and ${E}$ is a measurable set with ${\mu(E) \leq \delta}$.

The dominated convergence theorem does not have an analogue in the uniformly integrable setting:

Exercise 12 Give an example of a sequence ${f_n}$ of uniformly integrable functions that converge pointwise almost everywhere to zero, but do not converge almost uniformly, in measure, or in ${L^1}$ norm.

However, one does have an analogue of Exercise 7:

Theorem 5 (Uniformly integrable convergence in measure) Let ${f_n: X \rightarrow {\bf C}}$ be a uniformly integrable sequence of functions, and let ${f: X \rightarrow {\bf C}}$ be another function. Then ${f_n}$ converges in ${L^1}$ norm to ${f}$ if and only if ${f_n}$ converges to ${f}$ in measure.

Proof: The “only if” part follows from Exercise 2, so we establish the “if” part.

By uniform integrability, there exists a finite ${A > 0}$ such that

$\displaystyle \int_X |f_n|\ d\mu \leq A$

for all ${n}$. By Exercise 6, there is a subsequence of the ${f_n}$ that converges pointwise almost everywhere to ${f}$. Applying Fatou’s lemma, we conclude that

$\displaystyle \int_X |f|\ d\mu \leq A,$

thus ${f}$ is absolutely integrable.

Now let ${\epsilon > 0}$ be arbitrary. By uniform integrability, one can find ${\delta > 0}$ such that

$\displaystyle \int_{|f_n| \leq \delta} |f_n|\ d\mu \leq \epsilon \ \ \ \ \ (1)$

for all ${n}$. By dominated convergence, and decreasing ${\delta}$ if necessary, we may say the same for ${f}$, thus

$\displaystyle \int_{|f| \leq \delta} |f|\ d\mu \leq \epsilon. \ \ \ \ \ (2)$

Let ${0 < \kappa < \delta/2}$ be another small quantity (that can depend on ${A,\epsilon,\delta}$) that we will choose a bit later. From (1), (2) and the hypothesis ${\kappa < \delta/2}$ we have

$\displaystyle \int_{|f_n-f| < \kappa; |f| \leq \delta/2} |f_n|\ d\mu \leq \epsilon$

and

$\displaystyle \int_{|f_n-f| < \kappa; |f| \leq \delta/2} |f|\ d\mu \leq \epsilon$

and hence by the triangle inequality

$\displaystyle \int_{|f_n-f| < \kappa; |f| \leq \delta/2} |f-f_n|\ d\mu \leq 2\epsilon. \ \ \ \ \ (3)$

Finally, from Markov’s inequality(Exercise 34 of Notes 3) we have

$\displaystyle \mu( \{ x: |f(x)| > \delta/2 \} ) \leq \frac{A}{\delta/2}$

and thus

$\displaystyle \int_{|f_n-f| < \kappa; |f| > \delta/2} |f-f_n|\ d\mu \leq \frac{A}{\delta/2} \kappa.$

In particular, by shrinking ${\kappa}$ further if necessary we see that

$\displaystyle \int_{|f_n-f| < \kappa; |f| > \delta/2} |f-f_n|\ d\mu \leq \epsilon$

and hence by (3)

$\displaystyle \int_{|f_n-f| < \kappa} |f-f_n|\ d\mu \leq 3\epsilon \ \ \ \ \ (4)$

for all ${n}$.

Meanwhile, since ${f_n}$ converges in measure to ${f}$, we know that there exists an ${N}$ (depending on ${\kappa}$) such that

$\displaystyle \mu( |f_n(x)-f(x)| \geq \kappa ) \leq \kappa$

for all ${n \geq N}$. Applying Exercise 11, we conclude (making ${\kappa}$ smaller if necessary) that

$\displaystyle \int_{|f_n-f| \geq \kappa} |f_n|\ d\mu \leq \epsilon$

and

$\displaystyle \int_{|f_n-f| \geq \kappa} |f|\ d\mu \leq \epsilon$

and hence by the triangle inequality

$\displaystyle \int_{|f_n-f| \geq \kappa} |f-f_n|\ d\mu \leq 2\epsilon$

for all ${n \geq N}$. Combining this with (4) we conclude that

$\displaystyle \|f_n-f\|_{L^1(\mu)} = \int_{X} |f-f_n|\ d\mu \leq 5\epsilon$

for all ${n \geq N}$, and so ${f_n}$ converges to ${f}$ in ${L^1}$ norm as desired. $\Box$

Finally, we recall two results from the previous notes for unsigned functions.

Exercise 13 (Monotone convergence theorem) Suppose that ${f_n: X \rightarrow [0,+\infty)}$ are measurable, monotone non-decreasing in ${n}$ and are such that ${\sup_n \int_X f_n\ d\mu < \infty}$. Show that ${f_n}$ converges in ${L^1}$ norm to ${\sup_n f_n}$. (Note that ${\sup_n f_n}$ can be infinite on a null set, but the definition of ${L^1}$ convergence can be easily modified to accomodate this.)

Exercise 14 (Defect version of Fatou’s lemma) Suppose that ${f_n: X \rightarrow [0,+\infty)}$ are measurable, are such that ${\sup_n \int_X f_n\ d\mu < \infty}$, and converge pointwise almost everywhere to some measurable limit ${f: X \rightarrow [0,+\infty)}$. Show that ${f_n}$ converges in ${L^1}$ norm to ${f}$ if and only if ${\int_X f_n\ d\mu}$ converges to ${\int_X f\ d\mu}$. Informally, we see that in the unsigned, bounded mass case, pointwise convergence implies ${L^1}$ norm convergence if and only if there is no loss of mass.

[The following material was added after the notes were first posted, and will eventually be moved to a more appropriate location.]

Exercise 15 Let ${(X,{\mathcal B},\mu)}$ be a measure space, let ${f_n: X \rightarrow {\bf C}}$ be a sequence of measurable functions converging pointwise almost everywhere as ${n \rightarrow \infty}$ to a measurable limit ${f: X \rightarrow {\bf C}}$, and for each ${n}$, let ${f_{n,m}: X \rightarrow {\bf C}}$ be a sequence of measurable functions converging pointwise almost everywhere as ${m \rightarrow\infty}$ (keeping ${n}$ fixed) to ${f_n}$.

1. If ${\mu(X)}$ is finite, show that there exists a sequence ${m_1,m_2,\ldots}$ such that ${f_{n,m_n}}$ converges pointwise almost everywhere to ${f}$.
2. Show the same claim is true if, instead of assuming that ${\mu(X)}$ is finite, we merely assume that ${X}$ is ${\sigma}$-finite, i.e. it is the countable union of sets of finite measure.

(The claim can fail if ${X}$ is not ${\sigma}$-finite. A counterexample is if ${X = {\bf N}^{\bf N}}$ with counting measure, ${f_n}$ and ${f}$ are identically zero for all ${n \in {\bf N}}$, and ${f_{n,m}}$ is the indicator function the space of all sequences ${(a_i)_{i \in {\bf N}} \in {\bf N}^{\bf N}}$ with ${a_n \geq m}$.)

Exercise 16 Suppose that ${f_n: X \rightarrow {\bf C}}$ are a dominated sequence of measurable functions, and let ${f: X \rightarrow {\bf C}}$ be another measurable function. Show that ${f_n}$ converges pointwise almost everywhere to ${f}$ if and only if ${f_n}$ converges in almost uniformly to ${f}$.

Exercise 17 Let ${f_n: X \rightarrow {\bf C}}$ be a sequence of measurable functions, and let ${f: X \rightarrow {\bf C}}$ be another measurable function. Show that the following are equivalent:

1. ${f_n}$ converges in measure to ${f}$.
2. Every subsequence ${f_{n_j}}$ of the ${f_n}$ has a further subsequence ${f_{n_{j_i}}}$ that converges almost uniformly to ${f}$.

Exercise 18 Let ${X}$ be a probability space. Given any real-valued measurable function ${f: X \rightarrow {\bf R}}$, we define the cumulative distribution function ${F: {\bf R} \rightarrow [0,1]}$ of ${f}$ to be the function ${F(\lambda) := \mu( \{x \in X: f(x) \leq \lambda \} )}$. Given another sequence ${f_n: X \rightarrow {\bf R}}$ of real-valued measurable functions, we say that ${f_n}$ converges in distribution to ${f}$ if the cumulative distribution function ${F_n(\lambda)}$ of ${f_n}$ converges pointwise to the cumulative distribution function ${F(\lambda)}$ of ${f}$ at all ${\lambda \in {\bf R}}$ for which ${F}$ is continuous.

1. Show that if ${f_n}$ converges to ${f}$ in any of the seven senses discussed above (uniformly, essentially uniformly, almost uniformly pointwise, pointwise almost everywhere, in ${L^1}$, or in measure), then it converges in distribution to ${f}$.
2. Give an example in which ${f_n}$ converges to ${f}$ in distribution, but not in any of the above seven senses.
3. Show that convergence in distribution is not linear, in the sense that if ${f_n}$ converges to ${f}$ in distribution, and ${g_n}$ converges to ${g}$, then ${f_n+g_n}$ need not converge to ${f+g}$.
4. Show that a sequence ${f_n}$ can converge in distribution to two different limits ${f, g}$, which are not equal almost everywhere.

Convergence in distribution (not to be confused with convergence in the sense of distributions, which we will study later in this sequence) is commonly used in probability; but, as the above exercise demonstrates, it is quite a weak notion of convergence, lacking many of the properties of the modes of convergence discussed here.