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Let $m$ be a probability measure on $W \subseteq \mathbb{R}^m$, so that $m(W)=1$.

Consider a measurable function $f:W \rightarrow \mathbb{R}_{\geq 0}$.

Say if the following holds true.

$$ \lim_{M \rightarrow \infty} m\left( \{ w \in W : \ f(w) \geq M \} \right) = 0. $$

In other words, when $M \rightarrow \infty$, does the set $\{ w \in W : \ f(w) \geq M \}$ necessarily have measure $0$?

If not, provide an example of such $f$; and eventually provide weak conditions on $f$ under which the limit set has measure $0$.

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    Not necessarily assuming integrability or boundedness conditions on $f$?2012-12-02
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    No, we do not assume $f$ being integrable and/or bounded.2012-12-02

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The sets $S_M:=\{x\in W, f(x)\geqslant M\}$ are nested, the measure is finite and their intersection is empty.

More precisely, the sequence $\{S_M^c\}_{M=0}^{+\infty}$ is increasing to $W$. Writing $A_M:=S_M^c\setminus S_{M-1}^c$ and using $\sigma$-additivity of $m$, we get that $\mu(S_m^c)\to 1$.

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    Can you please explain in more details? Are you aiming at a particular known result? Thanks.2012-12-02
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    What do mean by $S_M^c$? The closure of $S_M$? Then at the end, why $m(S_M^c) \rightarrow 1$?2012-12-02
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    I meant the complement (in $W$).2012-12-02
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    I see: the measure of the complement goes to $1$.2012-12-02
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    Seems to me that this follows from the fact that we cannot have "$f(x) = \infty$" for all $x \in X \subseteq W$ with $m(X) > 0$. Is that right?2012-12-02
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    Yes (otherwise we can cheat taking a function which is always $+\infty$).2012-12-02
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    Ok. So, generalizing, if $f: W \rightarrow \mathbb{R}_{\geq 0} \cup \{\infty\}$, then we just need $m( \{w \in W: f(w) = \infty\} ) = 0$. Is that right?2012-12-02