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In general if I have that \begin{equation} \mathbb{E}(X)<\infty \end{equation} does this imply that $|X|<\infty$ a.s? An attempted proof:

Let $(\Omega,\mathbb{F},\mathbb{P})$ be a probability space and $K>0$ be a constant, then \begin{equation} \mathbb{E}(X)=\int_{\Omega}XdP\,=\int_{\{|X|\leq K \}}XdP\,+\int_{\{|X|>K\}}XdP\, <\infty \end{equation} Taking the limit as $K\rightarrow \infty$ \begin{equation} \int_{\{|X|>K\}}XdP\, \rightarrow 0 \end{equation} Struggling how to finish from here or posssibly this isn't true? Thanks!

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  • $\begingroup$ But what if $X$ takes on the value $\infty$ on a set of measure zero? $\endgroup$ – mathse May 11 '14 at 9:52
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    $\begingroup$ @mathse measure zero doesn't count ("$|X|<\infty$ a.s.") $\endgroup$ – Hagen von Eitzen May 11 '14 at 9:53
  • $\begingroup$ Yes, I overlooked the "almost surely". $\endgroup$ – mathse May 11 '14 at 10:26
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    $\begingroup$ How about using Markov's Inequality? $P(X=\infty)=P(\cap_{n=1}^{\infty} X\geq n)=\lim_{k} P(\cap_{n=1}^{k}X\geq n)=\lim_{k} P(X\geq k) \leq \lim_{k} \frac{E[x]}{k}=0$ $\endgroup$ – Lindon Nov 10 '14 at 17:38
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If not $|X|<\infty$ a.s., then there is a set of positive measure where $|X|=\infty$. The integral $$ \int_\Omega X\,\mathrm dP= \int_\Omega \max\{0,X\}\,\mathrm dP-\int_\Omega \max\{0,-X\}\,\mathrm dP$$ is defined only when at most one of the summands is infinite. If $X=+\infty$ on a set of positive measure, then $X=-\infty$ only on a zero-set, and then $\mathbb E(X)=+\infty$. Similarly, if $X=-\infty$ on a set of positive measure, then $\mathbb E(X)=-\infty$. Thus we have the following possibilities:

  • $|\mathbb E(X)|<\infty$ and $|X|<\infty$ a.s.
  • $\mathbb E(X)=+\infty$ and $X>-\infty$ a.s.
  • $\mathbb E(X)=-\infty$ and $X<\infty$ a.s.
  • $X$ is not integrable ($\mathbb E(X)$ does not exixts)

(So specifically, as you wrote $\mathbb E(X)<\infty$ without absolute value, it is possible to have $|X|=\infty$ with positive probability)

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  • $\begingroup$ I think you've answered my question but I have two side queries. Firstly why does $X=+\infty$ on a set of positive measure, imply $X=-\infty$ only on a zero-set. Secondly, at the risk of sounding stupid, I don't fully understand your final comment. Surely when $\mathbb{E}(X)<\infty$ it is just a real number and therefore it's absolute value is also finite? Assuming $X$ is integrable, $\mathbb{E}(X)<\infty$ corresponds to bullet points 1 and 3, in which cases $X<\infty$ a.s.? $\endgroup$ – Michael May 11 '14 at 10:30
  • $\begingroup$ Wait never mind the second point, I see that $\mathbb{E}(X)=-\infty$ then with positive probability $X=-\infty$ which gives with positive probability $|X|=\infty$. $\endgroup$ – Michael May 11 '14 at 10:42

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