# Lebesgue measure: $\mu(f-g(D))=0\Rightarrow\mu(f(D))=\mu(g(D))$

Let $f,g:\mathbb{R}\rightarrow\mathbb{R}$ be Lebesgue measurable functions, and let $D$ be a subset of $\mathbb{R}$. If $f(D)$ and $g(D)$ are measurable subsets of $\mathbb{R}$, and $(f-g)(D):=\{f(x)-g(x):x\in D\}$ is a measurable subset of $\mathbb{R}$ with measure $0$, is it necessarily true that $\mu(f(D))=\mu(g(D))$ (where $\mu$ denotes Lebesgue measure)?

This question was originally posed to me without the measure being specified as Lebesgue, and I was able to find a counter-example (using "unit mass" measure). However, for Lebesgue measure it appears to be true... it is clearly true if $f$ and $g$ are simple functions... although I don't see how a more general proof would go.

Thanks in advance for any help.

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No. Consider $f(x) = x$ and $g(x) = x - \lfloor x \rfloor$, the fractional part.