Any $L^1$ function is the subtraction between two pointwise limits of continuous functions

Question

Let $f\in L^1([0,1])$, and $\{\varphi_n\}_{n=1}^\infty$ and $\{\psi_n\}_{n=1}^\infty$ be two nondecreasing sequences of continuous functions on $[0,1]$. We want to prove that we can find two such sequences such that $\{\varphi_n(x)\}$ and $\{\psi_n(x)\}$ are bounded for a.e. $x\in [0,1]$, and moreover $$ f(x) = \lim_{n\to\infty} \varphi_n(x) - \lim_{n\to\infty}\psi_n(x).$$ How could we prove that such sequences exist? Thank you for your help!

Answer 1

First, take a sequence $\{f_k\}$ of continuous functions converging to $f$ in $L^1$. By taking a subsequence, we may assume $f_k \to f$ almost everywhere, and by passing to a further subsequence, we may assume $\|f_k - f_{k-1}\|_{L^1} \le 2^{-k}$ for each $k$.

Now, define $\phi_k$ and $\psi_k$ inductively as follows: Let $\phi_1 = f_1$ and $\psi_1 = 0$. Given $\phi_k$ and $\psi_k$, define $\phi_{k+1} = \phi_k + (f_{k+1} - f_k)^+$ and $\psi_{k+1} = \psi_k + (f_{k+1} - f_k)^-$. Note that for each $k$, $\phi_k - \psi_k = f_k$, and the sequences $\{\phi_k(x)\}$ and $\{\psi_k(x)\}$ are increasing for each $x$.

Right now, we have $f = \lim (\phi_k - \psi_k)$ a.e., but we need $f = \lim \phi_k - \lim \psi_k$ a.e. The only thing left to check is that $\{\phi_k\}$ and $\{\psi_k\}$ are bounded a.e., for which it suffices to show that $\lim \phi_k$ and $\lim \psi_k$ are in $L^1$. By Fatou, it suffices to show that $\|\phi_k\|_{L^1}$ and $\|\psi_k\|_{L^1}$ are uniformly bounded. That's why we assumed $\|f_k - f_{k-1}\|_{L^1} \le 2^{-k}$; this allows us to compute $$\|\phi_k\| \le \|f_1\| + \|f_2 - f_1\| + \dots + \|f_k - f_{k-1}\| \le \|f_1\| + \frac{1}{2},$$ and similarly for $\psi_k$.