Real Analysis, Folland Proposition 2.11/Exercise 10 Measurable Functions

Question

Question

Proposition 2.11 (Exercise 10) - The following implications are valid if and only if the measure is complete:

a.) If $f$ is measurable and $f = g$ $\mu$-a.e., then $g$ is measurable.

b.) If $f_n$ is measurable for $n\in \mathbb{N}$ and $f_n\rightarrow f$ $\mu$-a.e., then $f$ is measurable.

Attempted Proof a.) We want to show that for every Borel set $B\subset \mathbb{R}$, $g^{-1}(B)$ is measurable. Suppose $\mu$ is complete, since $f = g$ $\mu$-a.e., there exists a measurable set $E$ such that $\mu(E) = 0$, in fact, for all $x\notin E$ $f(x) = g(x)$. Then $$g^{-1}(B) = (g^{-1}(B)\cap E)\cup(f^{-1}(B)\setminus E)$$ since $f$ is measurable we have that $f^{-1}(B)$ is measurable, and since $\mu$ is complete, we have $g^{-1}(B)\cap E$ is measurable. Thus $g^{-1}(B)$ is measurable.

Now suppose part a.) holds. Let $N\subset E$, where $E$ has measure zero. Let $f = 1_{E}$ and $g = 1_{N}$ then $f = g$ a.e., so $g$ is measurable. Thus, $g^{-1}(\{1\}) = N$ is measurable. Therefore $\mu$ is complete.

Attempted proof b.) We are given $f_n$ to be measurable for $n\in\mathbb{N}$, and $f_n\rightarrow f$ a.e., from proposition 2.7 we can let $$\hat{f} = \lim_{n\rightarrow \infty}\sup f_n$$ Since $f_n$ is measurable then so is $\hat{f}$. Thus given the fact that $f_n\rightarrow f$ e.e. then we have $\hat{f} = f$ a.e., so by part a.) $f$ is measurable.

I am not sure how to proceed with the converse of part b.). Any suggestions is greatly appreciated.

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Background Information

Proposition 2.7 - If $\{f_j\}$ is a sequence of $\overline{\mathbb{R}}$-valued measurable functions on $(X,M)$, then the functions \begin{align*} g_1(x) &= \sup_{j}f_j(x) \ \ \ \ \ &g_3(x) = \lim_{j\rightarrow \infty}\sup f_j(x)\\ g_2(x) &= \inf_{j}f_j(x) \ \ \ \ \ &g_4(x) = \lim_{j\rightarrow \infty}\inf f_j(x) \end{align*} are all measurable. If $f(x) = \lim_{j\rightarrow \infty}f_j(x)$ exists for every $x\in X$, then $f$ is measurable.

Answer 1

Your proof is essentially correct. The proof of the last part, that is (b. $\Rightarrow$ $\mu$ is complete), is very similar to the proof that (a. $\Rightarrow$ $\mu$ is complete).

Here is the proof in details.

Proposition 2.11 (Exercise 10) - The following implications are valid if and only if the measure $\mu$ is complete:

a.) If $f$ is measurable and $f = g$ $\mu$-a.e., then $g$ is measurable.

b.) If $f_n$ is measurable for $n\in \mathbb{N}$ and $f_n\rightarrow f$ $\mu$-a.e., then $f$ is measurable.

Proof:

($\mu$ is complete $\Rightarrow$ a.)

Suppose $\mu$ is complete. Suppose $f$ is measurable and $f = g$ $\mu$-a.e.. Since $f = g$ $\mu$-a.e., there exists a measurable set $E$ such that $\mu(E) = 0$ and, for all $x\notin E$, $f(x) = g(x)$.

Given any Borel set $B\subset \mathbb{R}$, we have $$g^{-1}(B)= (g^{-1}(B)\cap E) \cup (g^{-1}(B) \cap E^c)=(g^{-1}(B)\cap E)\cup(f^{-1}(B)\setminus E)$$

since $f$ is measurable we have that $f^{-1}(B)$ is measurable, since $E$ is measurable, $f^{-1}(B)\setminus E$ is measurable. Since $\mu$ is complete and $\mu(E)=0$, we have $g^{-1}(B)\cap E \subset E$ is measurable. Thus $g^{-1}(B)$ is measurable.

(a. $\Rightarrow$ $\mu$ is complete)

Suppose a. holds. Let $E$ be a measurable set such that $\mu(E)=0$ and let $A$ be any subset of $E$.

Take $f=0$ (the null function) and $g=\chi_A$ (the indicator function of $A$). We have that $f$ is measurable and $f=g$ $\mu$-a.e., so by a.), we have that $g$ is measurable. Since $g^{-1}(\{1\})=A$, we have that $A$ is measurable. So $\mu$ is complete.

($\mu$ is complete $\Rightarrow$ b.)

Suppose $\mu$ is complete. Suppose $f_n$ is measurable for $n\in\mathbb{N}$, and $f_n\rightarrow f$ $\mu$-a.e..

Let $$\hat{f} = \lim_{n\rightarrow \infty}\sup f_n$$ Since $f_n$ is measurable, by proposition 2.7 we have that $\hat{f}$ is measurable. Thus given the fact that $f_n\rightarrow f$ $\mu$-a.e., we have $\hat{f} = f$ $\mu$-a.e., so, since $\mu$ is complete, a.) holds and we can conclude that $f$ is measurable.

(b. $\Rightarrow$ $\mu$ is complete)

Suppose b. holds. Let $E$ be a measurable set such that $\mu(E)=0$ and let $A$ be any subset of $E$.

Take $f_n=0$, for all $n\in \mathbb{N}$ and $f=\chi_A$ (the indicator function of $A$). We have that$f_n$ is measurable for $n\in \mathbb{N}$ and $f_n\rightarrow f$ $\mu$-a.e., so by b.), we have that $f$ is measurable. Since $f^{-1}(\{1\})=A$, we have that $A$ is measurable. So $\mu$ is complete.

Answer 2

A small correction:

There may exist a little error in your proof on a. $\Rightarrow$ $\mu$ is complete.

The Reason is that if $A \subset E $ and $\mu(E) = 0$ then $A$ may not be in measurable sets $\mathcal{M}$ and maybe $\mu$ cannot be defined on $A$.

That is the core tricky part of the proof, so for such $A$ one can modify like $$ f = \chi_E ,\quad g = a\chi_A + b\chi_{E-A} ,\quad a\not=b, a,b\not=1 $$ so that $E = \{f\not=g\}$ is $\mu$-measurable with zero measure and $f$ is measurable($f=g$ $\mu$-a.e),

by a. $g$ is measurable then $g^{-1}(a) = A$ is measurable.

As for b. $\Rightarrow$ $\mu$ is complete. One can set all $f_n$ equal to $\chi_E$ and $f_n \to g$ $\mu$-a.e where $g$ is defined as below.

Repeat the deduction when assuming a., one gets the completeness of $\mu$.