Suppose a sequence $\{f_{n} \}$ of functions in $L^{2}[0, 1]$ converges in measure to $f$, and furthermore, assume there exists constant $K$ such that $||f_{n}|| \leq K$ for all $n$. Show that $\{f_{n}\}$ converges weakly to $f$.
I'm following a proof which makes sense up to a point, but then there's a leap I don't understand. First, let $\{f_{n_{\ell}} \}$ be any subsequence of $\{f_{n}\}$. Then it itself has a subsequence $\{f_{n_{\ell_{j}}} \}$ that converges pointwise to $f$ on [0, 1].
First, we show $f$ is in $L^{2}[0, 1]$ by Fatou's Lemma:
$\int_{[0,1]} |f|^{2} \leq lim inf \int_{[0, 1]} |f_{n_{\ell_{j}}}|^{2} = lim inf (||f_{n_{\ell_{j}}}||_{2})^{2} \leq K^{2} < \infty$
Now, let $\epsilon > 0$, $g \in L^{2}[0, 1]$. Since $g^{2}$ is integrable, there exists $\delta > 0$ such that whenever $m(A) < \delta$, $\int_{A} g^{2} \leq (\epsilon/K)^{2}$.
Then, for all $j$, applying our observations together with Holder's inequality for $f_{n_{\ell_{j}}}$ and $g \cdot \chi_{A}$, we choose $\delta((\epsilon/K)^{2}))$ such that $m(A) < \delta$, yielding $\int_{A} |f_{n_{\ell_{j}}} \cdot g| < \epsilon$.
Therefore, $\{f_{n_{\ell_{j}}}\cdot g \}$ is uniformly integrable, and it converges pointwise to $fg$ a.e. Thus, we can pass the limit under the integral sign by the Vitali Convergence Theorem, yielding
$lim \int_{[0, 1]} (f_{n_{\ell_{j}}} \cdot g) = \int_{[0, 1]} fg$.
Thus, we have that $\{\int_{[0, 1]} f_{n_{\ell_{j}}} g \}$ goes to $\int_{[0, 1]} fg$ for any $g \in L^{2}[0, 1]$.
From here I want to be able to conclude that every subsequence $\{\int_{[0, 1]} f_{n_{\ell}} g \}$ of $\{ \int_{[0, 1]} f_{n} g \}$ converges to $\int_{[0, 1]} fg$. How can I make this leap though? The entire process above works with reference to a particular sub-subsequence $\{f_{n_{\ell_{j}}} \}$.
If $\{a_n\}$ is a sequence of real numbers and every subsequence of it has a further subsequence converging to $a$ then $a_n \to a$. You can prove this by contradiction. Apply this criterion with $g$ fixed.