$ \lim\limits_{p\to +0}\int_X |f|^p d\mu = \mu(\{ x\in X | f(x) \neq 0\} $

Question

Measure space $(X,\mathcal{M},\mu)$, $f$ is integrable on $X$. Prove

$ \lim\limits_{p\to +0}\int_X |f|^p d\mu = \mu(\{ x\in X | f(x) \neq 0\} $

Answer 1

Note that $|f|^p(x)$ behaves differently depending on whether $f(x) = 0$ or not. If $f(x) = 0$, then $|f|^p(x) = 0$, always, so we may without loss of generality restrict to the portion of $X$ where $f(x) \neq 0$; call it $Y$. Conversely, if $x \neq 0$, then $|f|^p(x) \to 1$ as $p \to 0^+$. Hence, you can use dominated convergence theorem to prove the claim. Partition $Y = X_1 \cup X_2$ where $X_1 = \{x : 0 < |f|(x) \leq 1\}$ and $X_2 = \{x : |f|(x) > 1\}$. On $X_1$, $|f|^p$ is bounded by $1$ which is integrable provided that $\mu(X_1) < + \infty$ (if $\mu(X_1) = +\infty$, you need a separate reasoning, but it's simple), hence dominated convergence theorem can be applied. Likewise, on $X_2$, $|f|^p$ is bounded by $|f|$, which is integrable by assumption, hence again the theorem can be applied. Working on each space separately, you find $$\lim_{p \to 0+} \int_{Y} |f|^p d\mu = \int_{Y} \lim_{p \to 0+} |f|^p d\mu = \int_{Y} 1 \ d \mu = \mu(Y)$$

Answer 2

Hint:

You need to think of a justification to be able to bring the limit inside the integral. I would start looking at the dominated convergence theorem and look for a way to construct a sequence of functions out of the $|f|^p$ that is bounded above.

I would also look into the types of functions that are in $L^p$ for all p.