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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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    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...2012-04-24

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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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    @Adam: Typo. ${}$2012-05-02