# Real Analysis, Folland Theorem 2.14 (Monotone Convergence Theorem)

Theorem 2.14 (MCT) - If $\{f_n\}$ is a sequence in $L^+$ such that $f_{n}\leq f_{n+1}$ for all $n$, and $f = \lim_{n\rightarrow \infty}f_n (=\sup_n f_n)$, then $\int f = \lim_{n\rightarrow \infty}\int f_n$.

Attempted proof - Let $f = \lim_{n\rightarrow \infty}f_n (= \sup_{n}f_n)$ then $\int f \geq \int f_n$ for all $n$. So $$\int f \geq \lim_{n\rightarrow \infty}\int f_n$$ Let $E_n = \{f_n\geq \alpha f\}$ where $\alpha \in (0,1)$ and $E_n$ is measurable. Further, let $\phi = \sum_{n}a_n\chi_{E_n}$ such that $\phi\leq f$ then we have a finite disjoint family $\{E_n\}_{n}$ such that $X = \bigcup_{n}E_n$. Also, $E_1\subset E_2\subset \ldots$ since $f_1\leq f_2\leq \ldots$. Since $f_n\rightarrow f$ then $$\int f_n\geq \int_{E_n}f_n\geq \int_{E_n}\alpha\phi = \alpha\int_{E_n}\phi$$ Taking the limits we get $$\lim_{n\rightarrow \infty}\int f_n\geq \lim_{n\rightarrow \infty}\alpha \int_{E_n}\phi = \alpha\int \phi$$ Since $\phi$ is arbitrary we have $$\lim_{n\rightarrow \infty}\int f_n\geq \alpha\int f$$ Since $\alpha$ is arbitrary $$\lim_{n\rightarrow \infty}\int f_n\geq \int f$$ Therefore $$\int f = \lim_{n\rightarrow \infty}\int f_n$$

I am not sure if this is right or if there is an easier way of proving MCT. Any suggestions is greatly appreciated.

• This is a good outline, but I think it's missing a bit. For starters, you have to show that $f$ is measurable. Then, I think you need to show that for any simple function $\theta\leq f$ there is a $\phi$ of the form you described such that $\theta\leq\phi$. – Callus Jun 23 '16 at 21:29
• You'll need to use the fact that $E_n$ is a decreasing sequence and you can switch limits with the integral over $E_n$. – Alex R. Jun 23 '16 at 21:33

## 1 Answer

@Wolfy your proof is correct. Just some minor adjustments are needed. I Have copied your proof, making such adjustments.

Theorem 2.14 (MCT) - If $\{f_n\}$ is a sequence in $L^+$ such that $f_{n}\leq f_{n+1}$ for all $n$, and $f = \lim_{n\rightarrow \infty}f_n (=\sup_n f_n)$, then $\int f = \lim_{n\rightarrow \infty}\int f_n$.

Proof - Let $f = \lim_{n\rightarrow \infty}f_n (= \sup_{n}f_n)$ then $\int f \geq \int f_n$ for all $n$. So $$\int f \geq \lim_{n\rightarrow \infty}\int f_n$$ Let $\phi$ be any simple function such that $0\leq \phi \leq f$ and
let $E_n = \{f_n\geq \alpha \phi\}$ where $\alpha \in (0,1)$ and $E_n$ is measurable. Further, then we have an non-decreasing family $\{E_n\}_{n}$ such that $X = \bigcup_{n}E_n$. So, we have $$\int f_n\geq \int_{E_n}f_n\geq \int_{E_n}\alpha\phi = \alpha\int_{E_n}\phi$$ Taking the limits we get $$\lim_{n\rightarrow \infty}\int f_n\geq \lim_{n\rightarrow \infty}\alpha \int_{E_n}\phi = \alpha\int \phi$$ Since $\phi$ is arbitrary, by the definition of $\int f$, we have $$\lim_{n\rightarrow \infty}\int f_n\geq \alpha\int f$$ Since $\alpha \ inin (0,1)$ is arbitrary, we have $$\lim_{n\rightarrow \infty}\int f_n\geq \int f$$ Therefore $$\int f = \lim_{n\rightarrow \infty}\int f_n$$