Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have an observation in my real analysis lecture notes that states that if $f,g:\,(X,\mu)\to\mathbb{R}$ (with Borel's -$\sigma$ algebra ) and $f=g$ almost everywhere then if $g$ is Lebesgue measurable then so if $f$.

I don't understand why this is true, can someone please explain ?

I tried looking at some Borel set and on it source, but I can't figure why if we change $g$ in some measure $0$ of points then the source is still in Lebesgue -$\sigma$ algebra

share|cite|improve this question
up vote 4 down vote accepted

Let $A$ be a measurable subset of $\mathbb R$. We must show that $Y=f^{-1}(A)$ is measurable in $X$.

By hypothesis, the set

$$ N=\lbrace x \in X | f(x) \neq g(x) \rbrace $$ is measurable and has measure $0$. We deduce that $Y \cap N$ is also measurable with measure $0$. So it suffices to show that $Y\setminus N$ is measurable.

Now $Y\setminus N =g^{-1}(A) \setminus N$ is a difference of two measurable sets, qed.

share|cite|improve this answer
Thanks, got it! – Belgi Dec 30 '12 at 7:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.