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A). Given that $ |X_n| \leq Y $ and $Y \in L$. Try to show $X_n$ is lebesgue integrable.

b). Try to give any example for which $X_n \longrightarrow^{L} X$ yet $\not\exists Y \in L$ with $|X_n| \leq Y$.

c). Try to give any example for which $X_n \to X$ w.p.1, and $X_n, X \in L$ yet $X_n \not\longrightarrow^L X$.

d). If $X_n$ is uniform integrable, does it follow that $g(X_n)$ is uniform integrable if g is continuous?

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    Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post.2012-10-28
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    Maybe $L$ denote the set of integrable random variables defined on a probability space. In b), you have to give more precisions about what $X_n\to X$ means (in which sense), even if we can guess it.2012-10-28
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    Dear David, I have rewrite my description. In part b) what I mean is $X_n \longrightarrow^L X$.2012-10-28

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  1. There is no need of the parameter $n$: if $0\leq X\leq Y$ and $Y$ is integrable, it follows from the definition of Lebesgue integral that $X$ is integrable. If $0\leq s\leq X$ is a simple function, then $0\leq s\leq Y$ so $\sup\{\int S,0\leq S\leq X,S\mbox{ simple}\}$ is finite.

  2. Try $X_n:=\sqrt n\chi_{((n+1)^{—1},n^{-1})}$.

  3. Try $X_n:=n\chi_{((n+1)^{—1},n^{-1})}$.

  4. Take $f$ an integrable function which is not in $L^2$. Then $X_n(x):=f(x+n)$ and $g(x)=x^2$.

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    Thanks for your answer, David. I have two wonderings: Is $\chi$ in part 2 means chi-square density? And how to show g(x) is not uniform integrable in part 4?2012-10-28
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    No, it's the characteristic function of a set. You have to show that $\{f(\cdot +n)^2\}$ is not uniformly integrable. If it was, it would be bounded $L^1$, so each function would be in $L^1$.2012-10-28
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    hi, david, can you please be more specific about what does $\chi_{(n+1)^{-1}, n^{-1}}$ mean?2012-10-29
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    Sorry, parenthesis where missing. $\chi_A(x)$ is $1$ if $x\in A$ and $0$ otherwise.2012-10-29