$\lim_{p\rightarrow 0}\int_X|f|^p\,d\mu=\mu(\{x\in X\,|\,f(x)\ne 0\})$

Question

Let $(X,\mathscr{M},\mu)$ be a measure space and $f$ be an essentially bounded function, i.e. $f\in L^{\infty}$. How do I show that

$\lim_{p\rightarrow 0}\int_X|f|^p\,d\mu=\mu(\{x\in X\,|\,f(x)\ne 0\})$.

In the cases where $f\in L^{p_0}$ for some $0<p_0<\infty$, the monotone/dominated convergence theorems show that this is indeed the case. But I don't see how I should proceed when $f$ is in $L^{\infty}$. Any advice?

Answer 1

We may assume that $f$ is not in $L^p$ for any $0<p<\infty$. Then the LHS is just $\infty$ so we only have to check that $\mu(\{f\ne0\})=\infty$. But, if $\mu(\{f\ne0\})$ is finite, then $\Vert f\Vert_1\leq \Vert f\Vert_{\infty}\mu(\{f\ne0\})$ hence $f$ is in $L^1$, a contradiction.