# Cohn, Exercise 2.4.11, Intuition behind my solution?

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.

• How did you get $\int 2f$ on the RHS of the last inequality? – Bungo Aug 28 '18 at 20:52
• @Bungo, Well $\int g_n = \int f_n + \int f - \int |f_n - f|$, right? The assumption is that $\int f_n \to \int f$, agreed? So $\liminf \int g_n \leq \lim \int f_n + \int f + \liminf- \int |f_n -f| = \int 2f + \liminf - \int |f_n -f|$, agreed? – Drew Brady Aug 28 '18 at 21:29
• Ah right, looks fine. – Bungo Aug 28 '18 at 21:31
• The proof is correct. But yeah, pretty unintuitive, you're right. – amsmath Aug 28 '18 at 22:25

It's easy to verify that $$\min(x,y) = {x+y-|x-y|\over2}$$ so $$g_n=2\min(f,f_n)$$
• Good point. I wonder if we can make the proof go through directly with $\min(f, f_n)$. – Drew Brady Aug 28 '18 at 23:07