15

How do I prove this proposition from Royden's Real Analysis:

If $\mu$ is a complete measure and $f$ is a measurable function, then $f=g$ almost everywhere implies $g$ is measurable.

In proving this proposition, what differs from the proof of a proposition from the first chapters stating:

If $f$ is a measurable function $f=g$ almost everywhere then $g$ is measurable.

In particular, what has to be modified in the following proof:

Take $E=\lbrace x \in X | f(x) \neq g(x) \rbrace,$ which is measurable and has measure $0$. For a measurable set $A$ in the range of $g$, we show that the set $Y=g^{-1}(A)$ is measurable. Now, $Y \cap E$ has is measurable with measure $0$. Since $Y \setminus E = f^{-1}(A) \setminus E$ is a difference of two measurable sets, we are done.

user58191
  • 497
  • 1
    You don't know that $E$ is measurable. All you know is that $E$ is a null-set, which by definition means that $E\subseteq F$, for some measurable null-set $F$, i.e. $F$ is measurable and $\mu(F)=0$. As Ilya points out: you cannot conclude that $Y\cap E$ is a measurable null-set, only that it is a null-set. – Stefan Hansen Jan 18 '13 at 09:55
  • Now, I'm lost on how to go about proving this. – user58191 Jan 18 '13 at 10:16
  • @StefanHansen Since the measure $\mu$ is complete, any subset of a null set is measurable; in particular, $E$ is measurable. Or am I missing something obvious? – saz Jan 30 '15 at 15:40
  • @saz: Yeah, you're right. I don't know what I was thinking of at the time. – Stefan Hansen Jan 30 '15 at 15:54
  • How can we prove the proposition from the first chapters "If f is a measurable function f=g almost everywhere then g is measurable" without knowing the measure is complete? I checked Royden's book but it only says subsets of E is measurable since mE = 0. – Ninja Oct 21 '20 at 19:33

2 Answers2

23

To fix notation: Suppose that $(X,\mathcal{A},\mu)$ is a complete measure space, $(Y,\mathcal{B})$ is a measure space and that $f:X \to Y$ is measurable. We have to show that

$$g^{-1}(B) \in \mathcal{A}$$

for all $B \in \mathcal{B}$. Now

$$\begin{align*} g^{-1}(B) &= \{x; g(x) \in B \} \\ &= \underbrace{\{x; g(x) \in B, f(x)=g(x)\}}_{=:A} \cup \underbrace{\{x; g(x) \in B, g(x) \neq f(x)\}}_{=:N}. \end{align*}$$

By definition, we have

$$N \subseteq \{x; f(x) \neq g(x)\}.$$

Since $f=g$ almost everywhere, this shows that $N$ is a subset of a nullset; hence measurable as $\mu$ is complete. Moreover,

$$\begin{align*} A &= \{x; g(x) \in B, f(x)=g(x)\} \\ &= \{x; f(x) \in B, f(x)=g(x)\} \\ &= \{x; f(x) \in B\} \cap \{x; f(x)=g(x)\} \end{align*}$$

is measurable as the intersection of two measurable sets (the first set at the right-hand side is measurable because $f$ is measurable and for the second one we note that

$$\{x; f(x)=g(x)\} = X \setminus \{x; f(x) \neq g(x)\}$$

is the complement of a measurable set and therefore measurable.)

saz
  • 123,507
  • I'm sorry, I should have mentioned that I'm looking for a proof of the more general case, $f:X\to Y$, where $Y$ is not necessarily the real line. (Just for the record, I found this elegant proof http://www.math.tamu.edu/~thomas.schlumprecht/hw6_math607_13c_sol.pdf , problem 1a, for the case $Y=\mathbb R$) – fonini Jan 30 '15 at 20:18
  • @fonini The proof works exactly the same way for the more general case. See my edited answer... – saz Jan 30 '15 at 20:19
  • Well, that was pretty simple indeed. Thanks :) – fonini Jan 30 '15 at 20:26
  • saz, could you explain why by definition $N \subseteq {x; f(x) \neq g(x)}$ ? which definition? –  May 02 '19 at 17:22
  • 1
    @Isa By the definition of $N$, we have $$N := {x; g(x) \in B, f(x) \neq g(x)}$$ and so $$N = {x; g(x) \in B} \cap {x; f(x) \neq g(x)} \subseteq {x; f(x) \neq g(x)}.$$ – saz May 02 '19 at 17:31
  • I see. and why since $f=g$ almost everywhere, this shows that $N$ is a subset of a nullset? Is it because ${x; f(x) \neq g(x)}$ has measure $0$? –  May 02 '19 at 17:38
  • 1
    @Isa $f=g$ almost everywhere implies that ${x; f(x) \neq g(x)}$ is a subset of a null set, i.e. ${x; f(x) \neq g(x)} \subseteq M$ for some null set $M$. This implies $N \subseteq {x; f(x) \neq g(x)} \subseteq M$, i.e. $N$ is subset of a null set. – saz May 02 '19 at 17:49
  • ok thank you! :) I've just notice that null set is a set such that $\mu(S)=0$ –  May 02 '19 at 17:59
4

Please, check assumptions of the proposition from the first chapters. When you are saying that $Y\cap E$ is measurable - how do you know this? You are not given that $Y$ is measurable, you have to prove it. But in case the measure is complete, by the definition of completeness if follows that $$ Y\cap E \subset E\text{ and }\mu(E) = 0\quad \Rightarrow \quad Y\cap E \text{ is measurable}. $$ Without completeness you can only conclude that $Y\cap E$ is $\mu$-null which does not imply measurability in general.

SBF
  • 36,568
  • I wasn't claiming $Y$ was measurable right away, I thought it's implied by the last two sentences.

    I guess $Y \cap E$ being measurable comes from the completeness property, is that the only gap from my attempt at proof?

     – user58191 Jan 18 '13 at 10:21
  • @user58191: I rather meant, that the measurabiltiy of $Y\cap E$ could be concluded either from measurability of $Y$ and the one of $E$, or using the fact that it is $\mu$-null and $\mu$ is complete. What I wrote is that you can't go the first way. Anyway, please tell me if you are still confused - I'll elaborate on unclear moments – SBF Jan 18 '13 at 10:24
  • It's a little clearer but I'm still a bit confused. So do I just add in the bit about $\mu$-completeness and $\mu$-nullity implying $Y \cap E$ is measurable (as above)? – user58191 Jan 18 '13 at 10:32
  • I meant, is that the only gap in what I have written above? – user58191 Jan 18 '13 at 10:53
  • @user58191: yes, the only part missing is the measurability of $Y\cap E$ which is justified due to the completeness of $\mu$ – SBF Jan 18 '13 at 11:37
  • 1
    Thanks! I think I get it already! – user58191 Jan 18 '13 at 12:08
  • @user58191: nice to know! – SBF Jan 18 '13 at 12:13