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

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?

• If $f \in L^1$, you can have the conclusion by using DCT. If $f \not \in L^1$, you can use Fatou's lemma to conclude. Nov 2, 2021 at 7:00
• Question about the "direct" solution with DCT, I don't know. Nov 2, 2021 at 7:01
• if you write $\int f$ don't you mean $f$ is integrable Nov 2, 2021 at 7:05
• @AdityaDwivedi The integral may be infinite
– John
Nov 2, 2021 at 7:34
• @John: Just direct, $\int \liminf g_n d \mu \le \liminf \int g_n d\mu$ Nov 2, 2021 at 8:22

## 1 Answer

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