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Monotone Convergence Theorem- If $\{f_n\}$ is a sequence in $L^{+}$ such that $f_j\leq f_{j+1}$ for all $j$, and $f = \lim_{n\rightarrow \infty}f_n(=\sup_{n}f_n)$, then $\int f = \lim_{n\rightarrow\infty}\int f_n$

Fatou's Lemma - If $\{f_n\}$ is any sequence in $L^{+}$, then $$\int f = \int(\lim\inf f_n)\leq \lim\inf\int f_n$$

Let $\{f_n\}_{n\in\mathbb{N}}\subset L^{+}$, then by Fatou's Lemma $$\int f = \int (\lim_{n\rightarrow \infty} f_n) \leq \lim_{n\rightarrow \infty}\inf\int f_n$$ we know that $f = \lim_{n\rightarrow \infty}f_n (=\sup_{n}f_n)$, and that $f_n\leq f$ hence $\int f_n \leq \int f$ for all $n\in \mathbb{N}$, it is clear then that $\lim_{n\rightarrow \infty}\sup\int f_n \leq \int f$. Therefore, $$\lim_{n\rightarrow \infty}\sup\int f_n\leq \int f = \int \lim_{n\rightarrow \infty}\inf f_n \leq \lim_{n\rightarrow \infty}\inf\int f_n = \lim_{n\rightarrow \infty}\int f_n$$ so, $\int f = \lim_{n\rightarrow \infty}\int f_n$

Just want to make sure this is correct, any suggestions is greatly appreciated.

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  • $\begingroup$ Yes, this is correct. $\endgroup$ – Moya Nov 14 '15 at 0:14
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Yes, your solution is correct. It only needs a minor adjustment regarding the indexes, to make it clearer.

Monotone Convergence Theorem- If $\{f_n\}$ is a sequence in $L^{+}$ such that $f_j\leq f_{j+1}$ for all $j$, and $f = \lim_{n\rightarrow \infty}f_n(=\sup_{n}f_n)$, then $\int f = \lim_{n\rightarrow\infty}\int f_n$

Fatou's Lemma - If $\{f_n\}$ is any sequence in $L^{+}$, then $$\int f = \int(\lim\inf f_n)\leq \lim\inf\int f_n$$

Let $\{f_n\}_{n\in\mathbb{N}}\subset L^{+}$, then by Fatou's Lemma $$\int f = \int (\lim_{n\rightarrow \infty} f_n) \leq \lim_{n\rightarrow \infty}\inf_{k\geq n}\int f_k$$ we know that $f = \lim_{n\rightarrow \infty}f_n (=\sup_{n}f_n)$, and that $f_n\leq f$ hence $\int f_n \leq \int f$ for all $n\in \mathbb{N}$, it is clear then that $\lim_{n\rightarrow \infty}\sup_{k\geq n}\int f_k \leq \int f$. Therefore, $$\limsup_n\int f_n=\lim_{n\rightarrow \infty}\sup_{k\geq n}\int f_k\leq \int f = \int \liminf_n f_n \leq \lim_{n\rightarrow \infty}\inf_{k\geq n}\int f_k = \liminf_n\int f_n \tag {1}$$ Since, we also know that $$\liminf_n\int f_n \leq \limsup_n\int f_n$$ From $(1)$ we get $$ \int f = \liminf_n\int f_n = \limsup_n\int f_n$$ which means, $\int f = \lim_{n\rightarrow \infty}\int f_n$

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  • $\begingroup$ There is a small typo at $(1)$: $\lim_{n\rightarrow \infty}\inf_{k\geq n}\int f_n$ must be $\lim_{n\rightarrow \infty}\inf_{k\geq n}\int f_k$. $\endgroup$ – Ninja Nov 9 '16 at 4:49
  • $\begingroup$ @Ninja Yes. Thanks for pointing it. I have corrected it. $\endgroup$ – Ramiro Nov 10 '16 at 0:05

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