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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](http://math.stackexchange.com/q/24413/5363)'s an example of a function $f: \mathbb{R} \to \mathbb{R}$ whose integral over any bounded interval is infinite.2012-04-24
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    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?2012-04-24
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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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