# Measure theory : Almost Everywhere

This question seems like an easy measure theory problem but I don't know how to proceed.

Suppose $f$ is defined a.e., and suppose $g$ and $h$ are defined on all $\mathbb R^n$, and $g=f$ a.e. and $h=f$ a.e. Then prove that:

$g$ is measurable $\iff$ $h$ is measurable.

I think this is true if and only if notation is trivial but how to prove?

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Is this homework? Or rather: are your 5 questions (so far) all parts of a homework? – Did May 6 '12 at 14:07

1. $f = g$ almost everywhere means that the set $A := \{x : g(x)\neq f(x)\}$ has measure $0$.
2. $f = h$ almost everywhere means that the set $B := \{x : h(x)\neq f(x)\}$ has measure $0$.
The problem then asks about the relationship between $g$ and $h$, so it is natural to consider the set $C := \{x : g(x)\neq h(x)\}$. Step 1 should be to answer the following questions: How does $C$ relate to $A$ and $B$? What is the measure of $C$?
Now suppose that $g$ is measurable. This means $g^{-1}(I)$ is measurable for every (possible infinite) interval $I$. You next want to answer the question: How does the set $h^{-1}(I)$ relate to $g^{-1}(I)$ and $C$? You want to conclude that $h^{-1}(I)$ is measurable as well.