If $f\in\mathcal{L}^1(\mu)$, prove that $\{x\in X: f(x)\neq 0\}$ is the countable union of sets with finite $\mu$-measure.

Question

Tonight I am working on the following problem:

Suppose $(X,\mathcal{S},\mu)$ is a measure space and $f\in\mathcal{L}^1(\mu)$. Prove that $$\{x\in X: f(x)\neq 0\}$$ is the countable union of sets with finite $\mu$-measure.

My first reaction was to write down the definition of what it means for $f$ to be in $\mathcal{L}^1(\mu)$:

Suppose $(X,\mathcal{S},\mu)$ is a measure space. An $\mathcal{S}$-measurable function $f:X\rightarrow\mathbf{R}$ is contained in $\mathcal{L}^1(\mu)$ if $$||f||_1=\int|f|\,d\mu<\infty.$$

Okay, so $\int|f|\,d\mu<\infty$. How does this help me? I'm not sure. After alot of searching, I read on what it means for a measure $\mu$ to be $\sigma$-finite. That got me thinking -- does $f\in\mathcal{L}^1(\mu)$ imply that $\mu$ is $\sigma$-finite? If so, then I think I could easily prove this using any of the conditions outlined in that Wikipedia page.

Am I on the right track? Any suggestions?

Thanks in advance!

Answer 1

This is a standard "layer cake" argument. Define $$E_n=\{x\in X\,|\, |f(x)|\ge \frac{1}{n}\}$$ and note that $\mu(E_n)$ must be finite, or $f$ would not be in $L^1$. But the union of all the $E_n$ is exactly the set where $f$ is nonzero.