3
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ This is great for me. Thank you! $\endgroup$ – Ergin Suer Oct 24 '14 at 18:28
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.