Finding limit of a convergent sequence

Question

Let $(X,\mathcal{A},\mu)$ be a measure space and let $f_n:X\to [0,+\infty]$ be a sequence of $\mathcal{A}$-measurable functions. For each $n\in \mathbb{N}$, we define an incresing sequence $(g_{n,k})$ of $[0,+\infty)$-valued simple $\mathcal{A}$-measurable functions such that $$f_n=\lim\limits_{k\to \infty}g_{n,k}(x)=f_n(x),$$ for all $x\in X$. For each $k\in \mathbb{N}$, we define $$h_k=\max(g_{1,k},\ldots, g_{k,k}).$$ Then $(h_k)$ is an increasing sequence of $[0,+\infty)$-valued simple $\mathcal{A}$-measurable functions. How can I show that $$\lim\limits_{k\to \infty}h_k(x)=\lim\limits_{n\to \infty}f_n(x)$$ for all $x\in X$?

Answer 1

By definition of $h_k$ we have $g_{n,k} \leq h_k$ for all $k \geq n$. Thus $$ f_n =\lim_{k \rightarrow \infty} g_{n,k} \leq \lim_{k \rightarrow \infty} h_k.$$ If $(f_n)_{n \in \mathbb{N}}$ a.e. pointwise convergent, we find $$\lim_{n \rightarrow \infty} f_n \leq \lim_{k \rightarrow \infty} h_k.$$ On the other hand, $(g_{n,k})_{k \in \mathbb{N}}$ is increasing and thus $g_{n,k} \leq f_n$. This implies $h_k \leq \max\{f_1,\ldots,f_k\}$. Now, we need that $(f_n)_{n \in \mathbb{N}}$ is also increasing (then the sequence is also convergent in $[0,\infty]$) to deduce $$\lim_{k \rightarrow \infty} h_k \leq \lim_{k \rightarrow \infty} f_k.$$

Otherwise this statement is not true, just take $f_1 = 1$ and $g_{1,k} =1$, then we always have $h_k \geq 2$ no matter how $f_n$, resp. $g_{n,k}$ is choosen for $n \geq 2$: Taking $f_n = g_{n,k} =0$ for $n \geq 2$, we get $\lim_{n \rightarrow \infty} f_n =0$, but $\lim_{k \rightarrow \infty} h_k =1$.