Cohn, Exercise 2.4.11, Intuition behind my solution?

Question

Here's the problem.

Let $(X, \mathscr{A}, \mu)$ be a measure space, and let $f,f_n$ be nonnegative functions that belong to $\mathscr{L}^1(X, \mathscr{A}, \mu, \mathbb{R})$. Prove that if $f_n \to f$ $\mu$-almost everywhere and $\int f_n \to \int f$, then $\int |f_n - f| \to 0$.

Note that $|f_n - f| \leq f_n + f$, so $g_n := f_n + f - |f_n -f| \geq 0$. Fatou gives $\int \liminf_n g_n \leq \liminf_n \int g_n$. Notice that $g_n \to 2f$, almost everywhere. Since the limit inferior is sub-additive, this means: $$ \int 2f = \int \liminf_n g_n \leq \liminf_n \int g_n \leq \int 2f + \liminf_n - \int |f_n - f| $$ Since $\liminf -a_n = -\limsup a_n$, we have $\limsup_n \int |f_n -f| \leq 0$, which implies the claim.

It took me a while to find this solution. Assuming it is correct, can someone help me understand why the definition of $g_n$ is natural in some sense here? I think this proof mechanically works, but isn't very intuitive to me.

Answer 1

It's easy to verify that $$\min(x,y) = {x+y-|x-y|\over2}$$ so $$g_n=2\min(f,f_n)$$

Answer 2

Looking back at this, and using the intuition from saulspatz, We can set $$ g_n = \min\{f_n, f\} $$ Since $g_n = \tfrac{f_n+ f - |f_n -f|}{2}$, $g_n \to f$, pointwise $\mu$-almost everywhere.
We then have $$ \int f = \int \liminf_n \min\{f_n, f\} \leq \liminf_n \int \min\{f_n, f\} $$ Therefore, $$ \limsup_n \int (f -f_n)_+ \leq 0. $$ Additionally, repeating the argument with $g_n = \min\{-f, -f_n\}$, $$ \limsup_n \int (f_n -f)_+ \leq 0. $$ Finally, $$ \limsup_n \int |f_n-f| \leq \limsup_n \int (f_n - f)_+ + \limsup_n \int (f- f_n)_+ \leq 0. $$ This proves the claim.