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I am reading Carother's section on "measurable functions" and I am stuck on a problem because I don't understand the notation.

What does it mean when we say $m(f(E))=0$.

I know the definition of $m^{*}(E)=inf\{\sum_{n=1}^{\infty}\ell(I_n) : E\subset \cup_{n=1}^{\infty}I_n\}$.

I also know that $m$ is defined to be the restriction of $m^{*}$ to the collection of measurable sets.

I know that $f:D\rightarrow \mathbb{R}$ is Lebesgue measureable if $D$ is measureable and if for each $\alpha \in \mathbb{R}$ the set $\{x \in D : f(x)>\alpha\}$ is measureable.

But what I don't know is what it means when we have $m(f(E))$.

I want to say it is $m(\{y : y=f(x)\in E\})$ but don't really understand what this means either. How do I put this together?

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    I assume that $f(E)$ in this case is the image of $E$ under the function $f$ - i.e. what $f$ maps $E$ into. He's saying that the measure of this set is zero. – 2017-02-11
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    $f(E)$ is a subset of $\R$ so if it is measurable, you can find its Lebesgue measure. However, it's not necessarily measurable.... – 2017-02-11
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    oh yeah I see. The problem actually says $f:[a,b] \rightarrow \mathbb{R}$ is continuous. That is important to understanding this I think. – 2017-02-11

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