Approximation of a $L^1$ function by a dominated sequence of continuous functions

Question

Consider $\mathbb{T}=\{z\in\mathbb{C}\mid |z|=1\}$ and the Lebesgue measure on it. Denote by $L^1(\mathbb{T})$ the set of integrable functions on $\mathbb{T}$ and by $C(\mathbb{T})$ the set of continuous functions on $\mathbb{T}$. I wonder whether or not the following proposition is true.

Proposition. Let $0\leq f\in L^1(\mathbb{T})$. Then there exist a sequence $\{g_n\}_{n\in\mathbb{N}}\subset C(\mathbb{T})$ and a function $g\in L^1(\mathbb{T})$ such that

1. $g_n\to f$ a.e.
2. $|g_n(t)|\leq g(t)$ a.e. for all $t\in\mathbb{T}$ and all $n\in\mathbb{N}$.

Just to add some context, I came up with this question while trying to find an alternative proof for a Fejér's-like theorem. I tried the usual techniques (density of continuous function in $L^1(\mathbb{T})$, Luzin's Theorem, etc) but I didn't succeed.

Answer 1

Yes, your proposition is true.

By density of the continuous functions, there is a sequence of continuous $g_n$ with $g_n \to f$ in $L^1$. Replacing $g_n$ with $g_n^+$, we can assume $g_n \ge 0$. Passing to a subsequence, we can also assume $g_n \to f$ almost everywhere. Passing to a further subsequence, we can assume $\|g_n - f\|_{L^1} \le 2^{-n}$.

(Actually, the first "pass to a subsequence" is unnecessary. As soon as $\sum_n \|g_n - f\|_{L^1} < \infty$, a Borel-Cantelli argument ensures that $g_n \to f$ almost everywhere.)

I claim $g_n$ is the desired sequence. It remains to construct the dominating function $g$. Let $h_n = (g_n - f)^+$. Then $h_n$ is measurable, $\|h_n\|_{L^1} \le \|g_n -f\|_{L^1} \le 2^{-n}$, and we have $h_n \ge 0$ and $g_n \le f + h_n$. Set $g = f + \sum_{n=1}^\infty h_n$. Now by monotone convergence $$\int \sum_{n=1}^\infty h_n = \sum_{n=1}^\infty \int h_n \le \sum_{n=1}^\infty 2^{-n} = 1$$ so $g$ is integrable. And for each $n$ we have $g_n \le f + h_n \le g$.

This argument didn't use any topology. Indeed, it still goes through if we replace $\mathbb{T}$ by any measure space $(X,\mu)$, and replace $C(\mathbb{T})$ by any dense subset $E \subset L^1(X,\mu)$ which is closed under the "positive part" operation. Maybe that latter condition can be weakened even further.

Answer 2

If $f$ is continuous, the result is a consequence of Arzela-Ascoli theorem. Because the sequence is non-negative, there is a sequence of measurable functions in $L^1$ that converges to $f$. Besides, thanks to the density of the continuous functions in $L^1$, we can assume that this functions are continuous. Let $F$ be such sequence of functions, i.e. $F=\{f_{n}:n\in \mathbb{N} \}⊂C(\mathbb{T})$. Any sub-sequence of functions in $F$ converges in $F\cup\{f\}⊂C(\mathbb{T})$. Arzela-Ascoli implies that $F$ is uniformly bounded, i.e. that the sequence satifies 2. For the general case, since $f$ is $L^1$, you can take a sequence of continuous functions converging to $f$. You can prove that there is a bound that holds for any case.