Question on mode of convergence of sequence of R.V.s

Question

I have two questions:

1) If $X_n$ is not integrable $X_n$ $\notin$ $L^1$. Does this mean that $X_n$ does not converge in $L^1$ for sure?

2) if $X_n(w)=\sqrt{n}I_{(0,1/n]}(w)$, I know that $X_n(w)$ converges in $L^1$, but if we want to use Lebesegue dominance theorem to claim that $X_n(w)$ converges in $L^1$, what is the R.V. $Y$ such that $|X_n|\leq Y$ $\forall n$? Is this $X_n(w)={n}I_{(0,1/n]}(w)$?

Answer 1

1) Typically $L^1$ convergence is not discussed for variables that are not themselves integrable. Yes, you can construct examples of non-integrable $(X_n)$ and $X$ where $E|X_n-X|\to 0$ (for example $X_n:=X$ where $X$ is not integrable), but such examples are not very interesting, and not much is gained from the knowledge that $L^1$ convergence occurs.

2) The dominating $Y$ needs to be free of $n$. You want a $Y$ that is at least as tall as each individual $X_n$. If you draw a picture of the $(X_n)$ you'll see a sequence of overlapping rectangles, having height $\sqrt n$ and centered over the interval $(0,\frac1n]$. You can convert these into non-overlapping rectangles where the $n$th rectangle has height $\sqrt n$ and is centered over the interval $(\frac 1{n+1},\frac 1n]$. This new representation of non-overlapping rectangles is a candidate for a dominating $Y$: $$ Y(\omega):=\sum_{n=1}^\infty \sqrt n I_{(\frac 1{n+1},\frac 1n]}(\omega)\tag1$$ To apply dominated convergence, your job is to confirm that each $X_n$ is indeed dominated by $Y$, and that this $Y$ is indeed integrable. (Since the indicators in (1) are non-overlapping, the integral of $Y$ equals the sum of the individual rectangle areas.)

Answer 2

1) Yes, it does not make usable sense to talk about convergence in $L^1$ of variables not in $L^1$. Of course it is possible that $\lVert X_n - X \rVert_1 \to 0$ even if $X_n$ and $X$ are not in $L^1$, but I would not call that "convergence in $L^1$".

2) Be careful: the domination in the dominated convergence theorem should NOT depend on $n$. Here a domination could not exist: you have convergence in $L^1$ but this is not by virtue of the dominated convergence theorem.