Theorem 1.26

Let $(f_{n})_{n\in\mathbb{N}}$ be a sequence of measurable functions, $f_{n}:X\to \mathbb{R}^{+}$ in measure space $(X,\mathcal{F},\mu)$. Assume $f_{n}\uparrow f$ pointwise i.e. $$0\le f_{0}\le f_{1}\le\dots\le f_{n}\le\dots f$$then, $$\lim_{ n \to \infty } \int\limits _{X}f_{n} \, d\mu=\int\limits _{X}f \, d\mu$$ # Proof First we note using the fact that limits of measurable functions are measurable that $f$ is measurable.

Then, since $$0\le f_{n}\le f_{n+1}$$ $\forall n$, then $$0\le\int\limits f_{n} \, d\mu\le\int\limits f_{n+1} \, d\mu$$since $\nu$ is a measure. Hence, $\left( \int\limits f_{n} \, d\mu \right)_{n\in\mathbb{N}}$ has a limit in $\bar{\mathbb{R}}$.

Now let $\alpha=\lim_{ n \to \infty }\left( \int\limits f_{n} \, d\mu \right)\in \bar{\mathbb{R}}$, we WTS $$\int\limits f \, d\mu =\alpha$$

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Since $\forall n\ge 1$, $0\le f_{n}\le f$ then $$\begin{gather} 0\le &\int\limits f_{n} \, d\mu &\le &\int\limits f \, d\mu \\ 0\le &\lim_{ n \to \infty } \int\limits f_{n} \, d\mu &\le &\int\limits f \, d\mu\\ 0\le &\alpha &\le &\int\limits f \, d\mu \end{gather}$$Now TS $\alpha\ge \int\limits f \, d\mu$ we instead show $$\forall 0<c<1:c\int\limits f \, d\mu \le \alpha$$