TITLE: Why does Dominated Convergence Theorem fail in this example?
QUESTION [6 upvotes]: Given a measure space, $(\Omega = [0,1], \mathcal{F} = \mathcal{B}[0,1], P)$, my professor said in class that the limit of the following expectation could not be evaluated using Dominating Convergence Theorem: for $n \ge 2$,
$$\lim_{n \to \infty} E\left(\cfrac{n}{\log n} \mathbb{1}_{[0,\frac{1}{n}]}(w)\right).
$$
But does $\cfrac{n}{\log n} \mathbb{1}_{[0, \frac{1}{n}]}(w) \to 0$ as $n \to \infty$? Then should it be bounded by some variable in $L_1(P)$ and so DCT is applicable? Or am I missing anything?

REPLY [3 votes]: Draw a picture of the sequence of functions! The $n$th function (call it $f_n$) is a box of height $h_n:=n/\log n$ sitting on the interval $[0, 1/n]$. As $n$ increases, the boxes move closer to the origin, getting thinner and taller. The point of this example is that the heights $h_n$ grow so quickly that any dominating function cannot be integrable.
To see this: Suppose function $g$ dominates the sequence $\{f_n\}$. Then $g$ 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 $g$ 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 $h_n=n/\log n$, so the series (1) diverges, which means $g$ is not integrable, hence you cannot satisfy the conditions of the Dominated Convergence theorem, even though $f_n\to0$ as $n\to\infty$ and $E(f_n)\to0$.
For other choices of $h_n$ the sum (1) will not diverge, and you've found a dominating $g$ that is integrable, and DCT will apply.
Note: The point of your prof's example is not that $h_n\to\infty$. If you try $h_n:=\sqrt n$ then $h_n\to\infty$ and yet DCT applies, since (1) converges for this choice of $h_n$.