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Let $\mu(\cdot)$ be a probability measure on $X$. Consider $f:X \rightarrow \mathbb{R}_{\geq 0}$.

Does Lebesgue measurability (w.r.t. $\mu(\cdot)$) of $f(\cdot)$ imply that $f(\cdot)$ is locally bounded?

If not, provide an example of measurable $f(\cdot)$ that is not locally bounded.

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Of course not... Here's an example of a function $f: \mathbb{R} \to \mathbb{R}$ whose integral over any bounded interval is infinite. – t.b. Apr 24 '12 at 21:41
Can you make your answer more clear? Is the fact that "the integral over any bounded interval is infinite" implying that the function is not locally bounded? Is the mentioned function measurable? – Adam Apr 24 '12 at 21:48
Yes I can: 1) Local boundedness implies that every point sits in an interval such that the integral over it is finite. 2) yes, of course the example is measurable, otherwise I wouldn't be able to speak of its integral. 3) Read the link before you ask... – t.b. Apr 24 '12 at 21:51

Recycle deleted answer:

Let $I_n$ be a countable basis of intervals for $\mathbb R$. Define $f_0 = 0$, then $f_{n + 1}$ is $f_n$ outside $I_{n + 1}$ and and $n + 1$ inside $I_{n + 1}$. So $(f_n)$ is an increasing sequence so $f = \sup f_n$ is measurable.

Now take an open interval $I$ and given $M > 0$ take a $n > M$ such that $I_n \subset I$. Now $\mu(|f| > M) \geq \mu(|f_n| > M) \geq |I_n| > 0$. So essentially unbounded on every open interval.

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How is $R_n$ defined? – Adam Apr 24 '12 at 22:06
@Adam: Typo. ${}$ – Jonas Teuwen May 2 '12 at 14:31

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