# $X$ integrable: Show $X\leq C$ a.e.

One little question:

If I have a probability space $(\Omega,\mathcal{U},P)$ and an integrable random variable $X$ on it, then does there exist a constant $c>0: \lvert X\rvert\leq c$ a.e.?

I know the following theorem for a measurable space $(M,\mathcal{A},\mu)$:

$f,g\colon M\to\overline{\mathbb{R}}$ and $$\int_A f\, dP\leq\int_A g\, d\mu~\forall A\in\mathcal{A}$$

then $f\leq g$ a.e..

I try to apply this on the situation here:

$X$ is integrable, i.e. $$\int_{\Omega}\lvert X\rvert\, dP<\infty,$$

i.e. there is a $c>0$, so that

$$\int_{\Omega}\lvert X\rvert\, dP\leq\int_{\Omega}c\, dP=c$$

So for $A=\Omega$, the theorem holds, because $X$ and $c$ are integrable and the inequation between the integrals is shown.

But does this hold for ALL $A\in\mathcal{U}$, and not only for $\Omega$?

Is $$\int_A\lvert X\rvert\, dP\leq\int_A c\, dP~\forall~A\in\mathcal{U}?$$

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## 1 Answer

Nope. Think about $X^{-1/2}$ on $(0,1)$.

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I found this theorem in a book! How can that be? –  math12 Dec 27 '13 at 18:48
@math12 I suggest you cite the book and exact wording and ask it as another question or an addendum to this one ;) –  AlexR Dec 27 '13 at 18:50