$L^1$ convergent subsequence of increasing functions uniformly bounded in $L^2$

Question

Problem: Suppose $f_n \colon [0,1] \to [0,\infty)$ for $n \in \mathbb{N}$ are increasing functions which are uniformly bounded in $L^2([0,1])$. Show that there exists a subsequence which converges in $L^1([0,1])$.

My attempt: If $M > 0$ is such that $\Vert f_n \Vert_2 \leq M$ for all $n \in \mathbb{N}$, then since the $f_n$ are increasing and nonnegative we have $$(1 - x)f_n(x)^2 = \int_x^1 f_n(x)^2 \,\mathrm{d}t\leq \int_x^1 \vert f_n(t) \vert^2 \,\mathrm{d}t \leq M^2$$ for all $n \in \mathbb{N}$, thus $$f_n(x) \leq g(x) = \frac{M}{\sqrt{1 - x}}$$ for all $n \in \mathbb{N}$. We note that $g \in L^1([0,1])$.

Suppose we can find an increasing function $f \colon [0,1] \to [0,\infty)$, a countable dense subset $S$ of $[0,1]$, and a subsequence $\{f_{n_k}\}_{k=1}^{\infty}$ such that $f_{n_k} \to f$ on $S$. Since $f$ is increasing, the set of discontinuity points $D$ is countable. If $x \in [0,1] \setminus D$, then there exists $\delta > 0$ such that $$\vert f(x) - f(y) \vert < \epsilon/6$$ if $\vert x - y \vert < \delta$. Since $S$ is dense, there exist $a,b \in S$ such that $$a < x < b, \quad b - a < \delta.$$ Then $$\vert f(b) - f(a) \vert \leq \vert f(b) - f(x) \vert + \vert f(x) - f(a) \vert < \epsilon/3.$$ Choose $N \in \mathbb{N}$ such that $$\vert f_{n_k}(y) - f(y) \vert < \epsilon/3$$ if $y = a,b$ and $n_k > N$. Since $\vert f(b) - f(a) \vert < \epsilon/3$ and $f,f_{n_k}$ are increasing, this implies that $$\vert f_{n_k}(x) - f(x) \vert < \epsilon$$ if $n_k > N$. This shows that $f_{n_k} \to f$ pointwise on $[0,1] \setminus D$, hence almost everywhere on $[0,1]$ since $D$ is countable. Since $f_{n_k} \leq g$ and $g \in L^1([0,1])$, the dominated convergence theorem guarantees that $f_{n_k} \to f$ in $L^1([0,1])$.

I am not 100% sure about some of the details of this, particularly how to choose such a function $f$ and such a subsequence. I think given any countable dense set $S$ which does not contain $1$, I can use a diagonalization argument to find a subsequence $\{f_{n_k}\}_{k=1}^{\infty}$ which converges pointwise on $S$ such that $\lim_{k\to\infty}f_{n_k}(x) \leq \lim_{k\to\infty}f_{n_k}(y)$ if $x < y$ with $x,y \in S$. I think then I can define $f$ on $[0,1)$ by $$f(x) = \begin{cases} \lim_{k\to\infty}f_{n_k}(x) & x \in S \\ \sup_{y \in S, y \leq x}\lim_{k\to\infty}f_{n_k}(y) & x \not\in S \end{cases}$$

Answer 1

Answering the correct question this time.

I. Local boundedness

Observe that for any $t>0$ $$\mu(f_n > t) \le \int_{\{f_n >t\}} f_n^2/t^2 \le \|f_n\|_2^2 /t^2\le M^2/t^2,$$ where $M=\sup \|f_n\|_2$.

II. Subsequence convergent on rationals

And since the functions are nondecreasing (in $x$) it follows that for any $t>M$, $f_n (x) \le t$ for all $x < 1-M^2/t^2$. A standard diagonal argument then shows that one can extract a subsequence $(n_k)$ such that $f_{n_k}(x)$ converges to a finite limit $f(x)$ as $k\to\infty$ at each rational $x\in [0,1)$.

III. Convergence a.e.

Extend the function $f$ to $[0,1)$ by letting $f(y) = \sup_{x\le y,x \in{\mathbb Q}}f(x)$. Note that this extension is nondecreasing, and is therefore continuous, possibly except on a countable set.

If $y\in (0,1)$ is a continuity point of $f$, then for $\epsilon>0$ there exist rationals $x_1<y < x_2$ with $0<f(x_2)-f(x_1)<\epsilon$. By taking $k$ large enough so that $|f_{n_k} (x_1) - f(x_1)|+|f_{n_k}(x_2)-f(x_2)|<\epsilon$, we have that

$$f(x)-f_{n_k}(x) \le f(x_2) - f_{n_k}(x_1)=f(x_2) -f(x_1) + f(x_1) - f_{n_k} (x_1)\le 2 \epsilon$$ Similarly, $$ f(x) - f_{n_k} (x) \ge f(x_1) - f_{n_k}(x_2) \ge \dots -2\epsilon.$$

Since $\epsilon$ is arbitrary, we proved that $f_{n_k}(x)\to f(x)$ at all continuity points of $f$ on $(0,1)$.

IV. Convergence in $L^1$

By Fatou, $f \in L^2$ with $\|f\|_2 \le M = \sup \|f_n\|_2$, and so by CS, $f \in L^1$. As noted in I, the functions $f_n$ are bounded above by $t$ on $1-M^2/t^2$. Therefore it follows from dominated convergence that for every $\epsilon>0$

Fix $\epsilon>0$. First observe that $$ \int_{[0,1-\epsilon]} |f_{n_k} -f| \to 0.$$

Next, observe that
$$ \int_{[1-\epsilon,1]} |f_{n_k} -f |\le \int_{[1-\epsilon,1]} |f_{n_k}| + \int_{[1-\epsilon,1]}|f|\le 2M \sqrt{\epsilon},$$

by CS. Combining both, $\limsup \int |f-f_{n_k}| \le 2 M \sqrt{\epsilon}$, completing the proof.

Answer 2

Let $f(x)=\lim_{n\to\infty} f_n(x)$ which exists (but may be $+\infty$) for all $x$ due to monotonicity. By Fatou's lemma,

$$ \int f \le \liminf \int f_n .$$ By C-S $\int |f_n| \le \sqrt{\int f_n^2}\times 1 =\|f_n\|_2$. As the norms on RHS are bounded by some constant, we have that $f\in L^1$.

Finally finish off with dominated convergence: $|f-f_n| = f-f_n \le f \in L^1$, and $|f_n - f|\to 0$ pointwise. Therefore, $\int |f-f_n| \to 0$. No need for subsequence.