For $f\in L^1(\mu)$, $\lim_{t\to \infty}(\mu\{x: |f(x)|>t\})=0$

Question

Let $(X,\Sigma,\mu)$ be a measure space and. $f\in L^1(\mu)$ show that: $\lim_{t\to \infty}\mu(\{x: |f(x)|>t\})=0$

I'm having some trouble with this proposition, do I use Fubini's theorem here?

Answer 1

No need for Fubini. Note \begin{align*}\mu(\{x\, : \, \lvert f(x) \rvert > t\}) &= \int_{\{x\, : \, \lvert f(x) \rvert > t\}} 1\, d\mu(x)\\ &\le \int_{\{x\, : \, \lvert f(x) \rvert > t\}} \frac{\lvert f(x) \rvert}{t} d\mu(x)\\ &\le \frac 1 t \int_X \lvert f(x) \rvert d\mu(x) = \frac 1 t \|f\|_{L^1(\mu)},\end{align*} where the first inequality holds since $\lvert f(x)\rvert > t \, \implies \, 1 \le \frac{\lvert f(x) \rvert}t$ on that set. Now since $\|f\|_{L^1(\mu)}$ is assumed to be finite, you can sent $t\to \infty$ to get the result.

Answer 2

Hint: if $E_t =\{x: |f(x)|>t\}$ then $$ \int_{E_t} |f| \ d\mu \ge t \ \mu(E_t).$$