Showing that the integral of $min(f, f_n)$ converges to the integral of $f$ where $f_n \to f$ a.e pointwise.

Question

Let $f_n\in L^{1}$ be a seq of nonnegative functions s.t $f_n \to f$ a.e ptwise. Show $\lim_{n \to \infty} \int (min(f, f_n) d\mu = \int f d\mu$.

For this, I wanted to use the dominated convergence theorem.

I have shown that $min(f, f_n)$ converges to $f$ pointwise since $|f - f_n| \ge |f - min(f, f_n)|$.

Moreover, $|min(f, f_n)| = min(f, f_n) \le f$. But I cannot use the dominated convergence theorem directly since the integral of $f$ may not be finite. Is there a way to fix this?

Answer 1

You will need more assumptions for DCT to work. For example suppose $\mu$ is the Lebesgue measure on $\mathbb{R}$. Now let $f_n = 1_{[-n,n]}$ be the indicator function for the set $[-n,n]$. $f_n$ is $L^1$ and $f_n \to f \equiv 1$ point wise. Hence DCT will not always work.

Instead you want to use the monotone convergence theorem. Specifically, define

$$ g_n(x) = \inf_{k \ge n} \min(f(x),f_k(x)). $$

Now $g_n$ is an increasing positive sequence and $g_n \to f$. So we have that

$$ \int g_nd\mu \to \int fd\mu. $$

Finally we see that for each $n$

$$ \int g_n d\mu \le \int \min(f,f_n)d\mu \le \int f d\mu$$

Thus, we have $\int \min(f,f_n)d\mu \to \int f d\mu$.