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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 $f$? – icurays1 Dec 2 '12 at 20:13
No, we do not assume $f$ being integrable and/or bounded. – Adam Dec 2 '12 at 20:14
up vote 3 down vote accepted

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. – Adam Dec 2 '12 at 20:19
What do mean by $S_M^c$? The closure of $S_M$? Then at the end, why $m(S_M^c) \rightarrow 1$? – Adam Dec 2 '12 at 20:30
I meant the complement (in $W$). – Davide Giraudo Dec 2 '12 at 20:31
I see: the measure of the complement goes to $1$. – Adam Dec 2 '12 at 20:32
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? – Adam Dec 2 '12 at 20:59

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