The Role Played by Sets of Measure Zero

1.35 Definition Let $P$ be a property which a point $x$ may or may not have. For instance, $P$ might be the property " $f(x) > 0$ " if $f$ is a given function, or it might be " $\{f_n(x)\}$ converges" if $\{f_n\}$ is a given sequence of functions.

If $\mu$ is a measure on a $\sigma$ -algebra $\mathfrak{M}$ and if $E \in \mathfrak{M}$ , the statement " $P$ holds almost everywhere on $E$ " (abbreviated to " $P$ holds a.e. on $E$ “) means that there exists an $N \in \mathfrak{M}$ such that $\mu(N) = 0$ , $N \subset E$ , and $P$ holds at every point of $E - N$ . This concept of a.e. depends of course very strongly on the given measure, and we shall write “a.e. $[\mu]$ " whenever clarity requires that the measure be indicated.

For example, if $f$ and $g$ are measurable functions and if

we say that $f = g$ a.e. $[\mu]$ on $X$ , and we may write $f \sim g$ . This is easily seen to be an equivalence relation. The transitivity ( $f \sim g$ and $g \sim h$ implies $f \sim h$ ) is a consequence of the fact that the union of two sets of measure 0 has measure 0.

Note that if $f \sim g$ , then, for every $E \in \mathfrak{M}$ ,

To see this, let $N$ be the set which appears in (1); then $E$ is the union of the disjoint sets $E - N$ and $E \cap N$ ; on $E - N$ , $f = g$ , and $\mu(E \cap N) = 0$ .

Thus, generally speaking, sets of measure 0 are negligible in integration. It ought to be true that every subset of a negligible set is negligible. But it may happen that some set $N \in \mathcal{M}$ with $\mu(N) = 0$ has a subset $E$ which is not a member of $\mathcal{M}$ . Of course we can define $\mu(E) = 0$ in this case. But will this extension of $\mu$ still be a measure, i.e., will it still be defined on a $\sigma$ -algebra? It is a pleasant fact that the answer is affirmative:

1.36 Theorem Let $(X, \mathcal{M}, \mu)$ be a measure space, let $\mathcal{M}^*$ be the collection of all $E \subset X$ for which there exist sets $A$ and $B \in \mathcal{M}$ such that $A \subset E \subset B$ and $\mu(B - A) = 0$ , and define $\mu(E) = \mu(A)$ in this situation. Then $\mathcal{M}^*$ is a $\sigma$ -algebra, and $\mu$ is a measure on $\mathcal{M}^*$ .

This extended measure $\mu$ is called complete, since all subsets of sets of measure 0 are now measurable; the $\sigma$ -algebra $\mathcal{M}^*$ is called the $\mu$ -completion of $\mathcal{M}$ . The theorem says that every measure can be completed, so, whenever it is convenient, we may assume that any given measure is complete; this just gives us more measurable sets, hence more measurable functions. Most measures that one meets in the ordinary course of events are already complete, but there are exceptions; one of these will occur in the proof of Fubini’s theorem in Chap. 8.

PROOF We begin by checking that $\mu$ is well defined for every $E \in \mathcal{M}^*$ . Suppose $A \subset E \subset B$ , $A_1 \subset E \subset B_1$ , and $\mu(B - A) = \mu(B_1 - A_1) = 0$ . (The letters $A$ and $B$ will denote members of $\mathcal{M}$ throughout this proof.) Since

we have $\mu(A - A_1) = 0$ , hence $\mu(A) = \mu(A \cap A_1)$ . For the same reason, $\mu(A_1) = \mu(A_1 \cap A)$ . We conclude that indeed $\mu(A_1) = \mu(A)$ .

Next, let us verify that $\mathcal{M}^*$ has the three defining properties of a $\sigma$ -algebra.

(i) $X \in \mathcal{M}^*$ , because $X \in \mathcal{M}$ and $\mathcal{M} \subset \mathcal{M}^*$ . (ii) If $A \subset E \subset B$ then $B^c \subset E^c \subset A^c$ . Thus $E \in \mathcal{M}^*$ implies $E^c \in \mathcal{M}^*$ , because $A^c - B^c = A^c \cap B = B - A$ . (iii) If $A_i \subset E_i \subset B_i$ , $E = \bigcup E_i$ , $A = \bigcup A_i$ , $B = \bigcup B_i$ , then $A \subset E \subset B$ and

Since countable unions of sets of measure zero have measure zero, it follows that $E \in \mathcal{M}^*$ if $E_i \in \mathcal{M}^*$ for $i = 1, 2, 3, \dots$ .

Finally, if the sets $E_i$ are disjoint in step (iii), the same is true of the sets $A_i$ , and we conclude that

This proves that $\mu$ is countably additive on $\mathcal{M}^*$ . ////

1.37 The fact that functions which are equal a.e. are indistinguishable as far as integration is concerned suggests that our definition of measurable function might profitably be enlarged. Let us call a function $f$ defined on a set $E \in \mathfrak{M}$ measurable on $X$ if $\mu(E^c) = 0$ and if $f^{-1}(V) \cap E$ is measurable for every open set $V$ . If we define $f(x) = 0$ for $x \in E^c$ , we obtain a measurable function on $X$ , in the old sense. If our measure happens to be complete, we can define $f$ on $E^c$ in a perfectly arbitrary manner, and we still get a measurable function. The integral of $f$ over any set $A \in \mathfrak{M}$ is independent of the definition of $f$ on $E^c$ ; therefore this definition need not even be specified at all.

There are many situations where this occurs naturally. For instance, a function $f$ on the real line may be differentiable only almost everywhere (with respect to Lebesgue measure), but under certain conditions it is still true that $f$ is the integral of its derivative; this will be discussed in Chap. 7. Or a sequence $\{f_n\}$ of measurable functions on $X$ may converge only almost everywhere; with our new definition of measurability, the limit is still a measurable function on $X$ , and we do not have to cut down to the set on which convergence actually occurs.

To illustrate, let us state a corollary of Lebesgue’s dominated convergence theorem in a form in which exceptional sets of measure zero are admitted:

1.38 Theorem Suppose $\{f_n\}$ is a sequence of complex measurable functions defined a.e. on $X$ such that

Then the series

converges for almost all $x$ , $f \in L^1(\mu)$ , and

PROOF Let $S_n$ be the set on which $f_n$ is defined, so that $\mu(S_n^c) = 0$ . Put $\varphi(x) = \sum |f_n(x)|$ , for $x \in S = \bigcap S_n$ . Then $\mu(S^c) = 0$ . By (1) and Theorem 1.27,

If $E = \{x \in S: \varphi(x) < \infty\}$ , it follows from (4) that $\mu(E^c) = 0$ . The series (2) converges absolutely for every $x \in E$ , and if $f(x)$ is defined by (2) for $x \in E$ , then $|f(x)| \le \varphi(x)$ on $E$ , so that $f \in L^1(\mu)$ on $E$ , by (4). If $g_n = f_1 + \cdots + f_n$ , then $|g_n| \le \varphi$ , $g_n(x) \to f(x)$ for all $x \in E$ , and Theorem 1.34 gives (3) with $E$ in place of $X$ . This is equivalent to (3), since $\mu(E^c) = 0$ . ////

Note that even if the $f_n$ were defined at every point of $X$ , (1) would only imply that (2) converges almost everywhere. Here are some other situations in which we can draw conclusions only almost everywhere:

1.39 Theorem(a) Suppose $f: X \to [0, \infty]$ is measurable, $E \in \mathcal{M}$ , and $\int_E f d\mu = 0$ . Then $f = 0$ a.e. on $E$ .

(b) Suppose $f \in L^1(\mu)$ and $\int_E f d\mu = 0$ for every $E \in \mathcal{M}$ . Then $f = 0$ a.e. on $X$ .

(c) Suppose $f \in L^1(\mu)$ and

Then there is a constant $\alpha$ such that $\alpha f = |f|$ a.e. on $X$ .

Note that (c) describes the condition under which equality holds in Theorem 1.33.

PROOF(a) If $A_n = \{x \in E: f(x) > 1/n\}$ , $n = 1, 2, 3, \dots$ , then

so that $\mu(A_n) = 0$ . Since $\{x \in E: f(x) > 0\} = \bigcup A_n$ , (a) follows.

(b) Put $f = u + iv$ , let $E = \{x: u(x) \ge 0\}$ . The real part of $\int_E f d\mu$ is then $\int_E u^+ d\mu$ . Hence $\int_E u^+ d\mu = 0$ , and (a) implies that $u^+ = 0$ a.e. We conclude similarly that

(c) Examine the proof of Theorem 1.33. Our present assumption implies that the last inequality in the proof of Theorem 1.33 must actually be an equality. Hence $\int_E (|f| - u) d\mu = 0$ . Since $|f| - u \ge 0$ , (a) shows that $|f| = u$ a.e. This says that the real part of $\alpha f$ is equal to $|\alpha f|$ a.e., hence $\alpha f = |\alpha f| = |f|$ a.e., which is the desired conclusion. ////

1.40 Theorem Suppose $\mu(X) < \infty$ , $f \in L^1(\mu)$ , $S$ is a closed set in the complex plane, and the averages

lie in $S$ for every $E \in \mathcal{M}$ with $\mu(E) > 0$ . Then $f(x) \in S$ for almost all $x \in X$ .

PROOF Let $\Delta$ be a closed circular disc (with center at $\alpha$ and radius $r > 0$ , say) in the complement of $S$ . Since $S^c$ is the union of countably many such discs, it is enough to prove that $\mu(E) = 0$ , where $E = f^{-1}(\Delta)$ .

If we had $\mu(E) > 0$ , then

which is impossible, since $A_E(f) \in S$ . Hence $\mu(E) = 0$ . ////

1.41 Theorem Let $\{E_k\}$ be a sequence of measurable sets in $X$ , such that

Then almost all $x \in X$ lie in at most finitely many of the sets $E_k$ .

PROOF If $A$ is the set of all $x$ which lie in infinitely many $E_k$ , we have to prove that $\mu(A) = 0$ . Put

For each $x$ , each term in this series is either 0 or 1. Hence $x \in A$ if and only if $g(x) = \infty$ . By Theorem 1.27, the integral of $g$ over $X$ is equal to the sum in (1). Thus $g \in L^1(\mu)$ , and so $g(x) < \infty$ a.e. ////