Monotone approximation of integrable functions

Question

It is well known that every $f \in L^1(0,1)$ (Lebesgue integrable functions on $(0,1)$) can be approximated by continuous (or even smooth) functions, in the sense that there exists a sequence of functions $(f_n)$ such that $f_n \in C^{\infty}[0,1]$ and $$ \lim_{n \to +\infty} f_n = f $$ almost surely (for almost every $x \in (0,1)$). I would like to know if something similar is possible if we want the sequence to be monotone in $n$? More precisely, my question is:

Let $f \in L^1(0,1)$. Does there exist a sequence $(f_n)$ of twice continuously differentiable functions $f_n \in C^2[0,1]$ such that

• $f_n(x) \ge f_{n+1}(x)$ for almost every $x \in (0,1)$,
• $f_n(x)$ converges to $f(x)$ for almost every $x \in (0,1)$?

Answer 1

This is not true. Let $f(x) = \ln x$, $f(0) = 0$. We have $f_1(0) = a$, and, as $f_1$ is continuous, $f_1(x) > a - 1$ for $x < \epsilon$ for some $\epsilon$. As $f_n$ is monotonic, it means $f_n(x) > a - 1$ for all $n$ and $x < \epsilon$.

But for $x < \min(\exp(a - 2), \epsilon)$ we have $\ln x < a - 2$, and thus $f_n(x) > \ln x + 1$ on $(0, \min(\exp(a - 2), \epsilon)$, so $f_n$ doesn't converge to $f$ a.e.

UPD: it's not possible even for simple $f$ and just continuous $f_n$.

Let $M$ be any measurable set sets s.t. for any interval $I$, $\mu(I) > \mu(M \cap I) > 0$ (see this answer how to construct it). Let $f(x) = \mathbb I_M(x)$.

Let $x_0$ be any point of $M$ s.t. $f_n(x) \to f(x)$ (such point exists, because $f_n \to f$ a.e., and $M$ has positive measure). Then, for some $n$ we have $f_n(x_0) > 1/2$, and, as $f$ is continuous, for some $\epsilon > 0$ we have $f_n(x) > 1/4$ for $x \in (x_0 - \epsilon, x_0 + \epsilon)$. But that means that $f_n$ doesn't converge to $0$ on set $N = (x_0 - \epsilon, x_0 + \epsilon) \setminus M$, which has positive measure, while $f$ is $0$ on this set.