# Why $g_n\to g$ a.e. if we set $g_n=\min\{g,f_n\}$?

Suppose $$\{f_n\}$$ is a sequence of (Lebesgue) measurable functions on $$\mathbb{R}^d$$ with each $$f_n\geq 0$$. If $$f_n(x)\to f(x)$$ for a.e. $$x$$, then by Fatou's lemma, we have $$\int f\leq\liminf_{n\to\infty}\int f_n.$$ Its proof begins with a non-negative function $$g$$ that is bounded and supported on a set $$E\subseteq\mathbb{R}^d$$ of finite measure with $$g\leq f$$. If we set $$g_n=\min\{g,f_n\}$$, then each $$g_n$$ is a measurable function and supported on $$E$$. Furthermore, $$g_n(x)\to g(x)$$ for a.e. $$x$$. By the bounded convergence theorem, ...

The above material is quoted from the book by Stein and Shakarchi with some minor changes, and I wonder why $$g_n\to g$$ almost everywhere. I wish I could offer you some useful ideas, but unfortunately I know nothing, which is why I'm here. I would much appreciate it if you could do me a favor. Thank you.

• Let $x$ be a point where convergence holds. Intuitively, $f_n(x)\simeq f(x)$ for large $n$. Since $f(x)\geq g(x)$, we might think $g_n(x)\simeq g(x)$ for large $n$.
– Boar
Jan 15, 2022 at 15:52
• The function $\mathbb{R}^2 \to \mathbb{R},$ $(y, z) \mapsto \min\{y, z\}$ is continuous. Therefore, for all $y \in \mathbb{R},$ the function $\mathbb{R} \to \mathbb{R},$ $z \mapsto \min\{y, z\}$ is continuous. Jan 15, 2022 at 17:51

Given $$x \in \mathbb{R^d}$$, we will look at two cases: (You may try writing out these arguments in a formal manner).

$$(i)$$ If $$f(x)=g(x)$$, then $$f_n(x)$$ converges to $$g(x)$$. This completes this case as $$g_n(x)=f_n(x)$$ or $$g_n(x)=g(x)$$ for each $$n\in \mathbb{N}$$.

$$(ii)$$ If $$f(x)>g(x)$$, then as $$f_n(x)$$ converges to $$f(x)$$, for large $$n$$, $$f_n(x)>g(x)$$, thus $$g_n(x)=g(x)$$. So, again, $$g_n(x)\rightarrow g(x)$$.

The reason is because the function $$\min : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$$ is continuous. This follows from writing $$\min(a, b) = \frac{a + b}{2} - \frac{|a - b|}{2}$$, or from a direct argument by cases ($$a < b$$, $$a = b$$).

Thus in your case, since $$f_n \to f$$ a.e., it follows that $$\min(g, f_n) \to \min(g, f) = g$$ a.e.

I think I found the answer, but I need someone to help confirm it. Let $$x$$ be a point such that $$f_n(x)\to f(x)$$. By using an $$\epsilon$$-$$\delta$$ argument, I'm going to prove $$g_n(x)\to g(x)$$.

There are two cases to consider: $$f(x)=g(x)$$ and $$f(x)>g(x)$$. If $$f(x)=g(x)$$, choose $$N\in\mathbb{N}$$ s.t. $$n\geq N\Rightarrow|f_n(x)-f(x)|<\epsilon$$. By construction, $$|g_n(x)-g(x)|<\epsilon$$ whenever $$n\geq N$$. Now consider the case $$f(x)>g(x)$$. This time we choose $$N\in\mathbb{N}$$ s.t. $$n\geq N\Rightarrow|f_n(x)-f(x)|. Then, as long as $$n\geq N$$, we must have $$|g_n(x)-g(x)|<\epsilon$$.

Is my proof correct? Thank you.