Why does Dominated Convergence Theorem fail in this example? 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?
 A: 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 $h_n$.
A: Let $X_n=\frac n{\log n}\mathsf 1_{\left[0,\frac1n\right]}$. Note that $x\mapsto \frac x{\log x}$ is monotone decreasing on $\left(0,\frac1n\right]$ (we can assume WLOG that $X_n(0) = 0$ since $\{0\}$ is a set of measure zero), so for each $n$, $$M_n:=\sup_{\omega\in\left[0,\frac1n \right]}|X_n| = \frac1{n\log n}.$$ If $Y$ dominates $\{X_n\}$, then $M_n\leqslant Y$ for all $n$, which implies that $Y$ is not integrable.
However, we may evaluate this limit directly: $$\mathbb E\left[ \frac n{\log n}\mathsf 1_{\left[0,\frac1n\right]}\right] = \frac n{\log n}\cdot\frac 1n = \frac1{\log n}\stackrel{n\to\infty}\longrightarrow0. $$
A: For the dominated convergence theorem to apply to a sequence of functions (or random variables in your context) $f_n$, there needs to be a function $g\in L^1$ so that $|f_n|<g$ for all $n$. Note that it is not enought that $g\geq|\lim_{n\to\infty} f_n|$. 
If $$f_n=\frac{n}{\log n}\mathbf 1_{[0,1/n]},$$ then it is true that $f_n\to 0$ almost everywhere, but no matter what $L^1$ function $g$ we choose to compare to $f_n$, we will have that $f_n(x)>g(x)$ for some values of $x$ (a set of positive measure) and $n$ sufficiently large. This is because for each $n$, we can choose $x$ sufficiently large so that $f_n(x)$ is very large on the interval $[0,1/n]$.
