Limit of a sequence of measurable function.

Question

We know that for a nonnegative measurable function $f$ defined from some measure space $(\Omega,\mathcal{F},\mu)$ to $\mathbb{R}$, the integration of $f$ is defined as: $$\int f= \sup\{\int g:g\ \text{is simple and nonnegative and } g\leq f\}$$ So we can find a sequence $\{\int g_n\}$ in the set at left hand side such that $\int g_n\rightarrow \int f$ as $n\rightarrow \infty$. Let $$h_n= \max\{g_j:1\leq j\leq n\}$$ Show that $h_n$ increases to $f$ a.e. $(\mu)$ as $n\rightarrow \infty$.

Answer 1

Note that $0 \le g_n(x) \le h_n(x) \le h_{n+1}(x) \le f(x)$ for all $n,x$.

Hence $0 \le \int g_n \le \int h_n \le \int f$.

Since $h_n$ is non decreasing, we can let $h (x) = \lim_n h_n(x)$.

The monotone convergence theorem shows that $\int h_n \to \int h$ and hence $\int h = \int f$ from which we have $\int (f-h) = 0$. Since $f(x)-h(x) \ge 0$, we get $f(x) = h(x)$ ae. $x$ [$\mu$].