# Can we use the dominated convergence theorem in this example?

Define a sequence $\{X_n\}$ of random variables on $\Omega=[0,1]$ by

$X_n(\omega)=\begin{cases}n,&\omega\in[1-\frac{1}{n}, 1] \\0,& \omega\in[0,1-\frac{1}{n}) \end{cases}$

My question concerns whether the dominated convergence theorem is suitable for analysis of expectations of this sequence. As $n\to\infty$, the sequence approaches $0$ at all points in $\Omega$ except at the end point $1$, where the sequence diverges. So does this mean that we cannot use the DCT, since we cannot find a r.v. that bounds the sequence at all points in $\Omega$?

• I suppose $\Omega$ has the uniform probability distribution. In that case you are right. In fact, if the DCT applied, we would get $0=1$ Commented Jan 24, 2018 at 22:00

Draw a picture of your sequence. The $n$th variable $X_n$ is a box of height $h_n:=n$ and width $\frac1n$ sitting on the interval $[1-1/n, 1]$. As $n$ increases, the boxes move closer to the point $\omega=1$, getting thinner and taller. But the heights $h_n$ grow so quickly that any dominating function cannot be integrable.
To see this: Suppose random variable $Y$ dominates the sequence $\{X_n\}$. Then $Y$ is at least as tall as the entire array of boxes. The entire array of boxes can be decomposed into a disjoint union of boxes, with height $h_n$ and width $\frac1n - \frac1{n+1}=\frac1{n(n+1)}$, and therefore the integral of this dominating $Y$ has to exceed the sum of the areas of these disjoint boxes, which is $$\sum_{n=2}^\infty \frac{h_n}{n(n+1)}.\tag1$$ In your example $h_n=n$, so the series (1) diverges, which means $Y$ is not integrable, hence you cannot satisfy the conditions of the Dominated Convergence theorem.