Prove that $\lim_{j\rightarrow\infty}\int_1^\infty\frac{f_j(x)}{x}dx=\int_1^\infty\frac{f(x)}{x}dx$ under these conditions... Question: Let $\{f_j\}_{j\in\mathbb{N}}$ be a sequence of Lebesgue measurable functions satisfying $$\sup_{j\in\mathbb{N}}\int_1^\infty f_j^2(x)dx\leq1$$ such that $f_j\rightarrow f$ pointwise a.e.  Prove that $$\lim_{j\rightarrow\infty}\int_1^\infty\frac{f_j(x)}{x}dx=\int_1^\infty\frac{f(x)}{x}dx$$
My Thoughts:  By our assumption, suppose $f_n\rightarrow f$ pointwise a.e. Let $K$ be the set such that $m(K)=0$ such that $\forall\epsilon>0, \forall x\in((1,\infty)-K)$, there exists $J$ such that $\forall j>J$, $|f_j(x)-f(x)|<\epsilon$.  Now, I feel like if I could use DCT, then maybe I could get this to work... since $\sup_{j\in\mathbb{N}}\int_1^\infty f_j^2(x)dx\leq1$, could we bound $\frac{f_j(x)}{x}$ by $\frac{f_j^2(x)}{x}$?  I don't suppose so... I suppose I am just not quite sure how to tie our assumption to what we are trying to prove using maybe a more measure theoretic technique?  Any thoughts, suggestions, etc. are appreciated!  Thank you.
 A: We first observe that, by the Fatou's Lemma,
$$ \int_{1}^{\infty} f(x)^2 \, \mathrm{d}x \leq \liminf_{j\to\infty} \int_{1}^{\infty} f_j(x)^2 \, \mathrm{d}x \leq 1. $$
In particular, we obtain
$$ \sup_{j\in\mathbb{N}} \int_{1}^{\infty} (f(x) - f_j(x))^2 \, \mathrm{d}x \leq 4. $$
Now we fix $\epsilon \in (0, 1)$ and use Egoroff's Theorem to find $E \subseteq [1, \epsilon^{-1}]$ such that
$$
\operatorname{Leb}([1,\epsilon^{-1}]\setminus E)<\epsilon
\qquad\text{and}\qquad
f_j \to f \text{ uniformly on } E.
$$
Writing $F_1 = [1,\epsilon^{-1}]\setminus E$ and $F_2 = (\epsilon^{-1},\infty)$ for simplicity, we have
\begin{align*}
\left| \int_{1}^{\infty} \frac{f_j(x)}{x} \, \mathrm{d}x - \int_{1}^{\infty} \frac{f(x)}{x} \, \mathrm{d}x \right|
\leq \int_{E} \frac{\left| f_j(x) - f(x) \right|}{x} \, \mathrm{d}x + \int_{F_1\cup F_2} \frac{\left| f_j(x) - f(x) \right|}{x} \, \mathrm{d}x.
\end{align*}
The first term is easily controlled by the choice of $E$. Indeed,
$$ \lim_{j\to\infty} \int_{E} \frac{\left| f_j(x) - f(x) \right|}{x} \, \mathrm{d}x = 0 $$
by the uniform convergence. Next, by the Cauchy-Schwarz inequality,
\begin{align*}
\int_{F_1\cup F_2} \frac{\left| f_j(x) - f(x) \right|}{x} \, \mathrm{d}x
&\leq \left( \int_{1}^{\infty} (f_j(x) - f(x))^2 \, \mathrm{d}x \right)^{1/2}\left( \int_{1}^{\infty} \frac{\mathbf{1}_{F_1\cup F_2}(x)}{x^2} \, \mathrm{d}x \right)^{1/2} \\
&\leq 2 \left( \operatorname{Leb}(F_1) + \epsilon \right)^{1/2} \\
&\leq \sqrt{8\epsilon}.
\end{align*}
Combining altogether,
$$ \limsup_{j\to\infty} \left| \int_{1}^{\infty} \frac{f_j(x)}{x} \, \mathrm{d}x - \int_{1}^{\infty} \frac{f(x)}{x} \, \mathrm{d}x \right| \leq \sqrt{8\epsilon}. $$
Since the left-hand side is independent of the choice of $\epsilon$, letting $\epsilon \downarrow 0$ proves the claim.

Remark. This is an adaptation of the proof of a more general result called Vitali Convergence Theorem.
