Let $(X,\Sigma , \mu)$ be a measure space. Show that if $f \ge 0$ and $\int fd\mu<\infty$, then for any $a>0$ the set $E_a:=\{f\ge a\}$ has finite $\mu$-measure.

My attempt: We know that there is sequence $\phi_n$ of simple functions with $lim_{n\to\infty}\phi_n=f$. By the monotone converges theorem $\int fd\mu=lim_{n\to\infty}\int \phi_nd\mu<\infty$.

How can I conclude the proof? I am very new in measure theory. Thanks!

  • $\begingroup$ What would happen if $\mu(E_1)=\infty$. $\endgroup$ – azarel Oct 24 '14 at 18:10
  • $\begingroup$ Intuitiviely, it is obvious but I don't know how to write it. May be I miss some basics $\endgroup$ – Ergin Suer Oct 24 '14 at 18:14

Here is how I would solve this problem, given $a > 0$:

We know $\mu(\{x \mid f(x) > a\}) = \int \limits_{\{x \mid f(x) > a\}} 1 \,d\mu < \int \limits_{\{x \mid f(x) > a\}} \frac{f(x)}{a} \,d\mu < \frac{1}{a}\int \limits_{X} f(x) \,d\mu < \infty$.

Do you see why each inequality holds?

In case not, or if anyone else may want to see the details, here they are:

$\mu(\{x \mid f(x) > a\}) = \int \limits_{\{x \mid f(x) > a\}} 1 \,d\mu = \int \limits_{X} 1 \cdot \chi_{\{x \mid f(x) > a\}} \,d\mu < \int \limits_{X} \frac{f(x)}{a}\chi_{\{x \mid f(x) > a\}} \,d\mu < \frac{1}{a}\int \limits_{X} f(x) \,d\mu < \infty$

  • $\begingroup$ This is great for me. Thank you! $\endgroup$ – Ergin Suer Oct 24 '14 at 18:28

For $a>0$ prescribe $f_a$ by $x\mapsto a$ if $f(x)\geq a$ and $x\mapsto 0$ otherwise.

Then $f_a$ is a measurable function with $0\leq f_a(x)\leq f(x)$ for each $x$.

Consequently: $$a\times\mu\left(\left\{ f\geq a\right\} \right)=\int f_{a}d\mu\leq\int fd\mu<\infty$$

hence:$$\mu\left(\left\{ f\geq a\right\} \right)<\infty$$


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