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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
    Commented Nov 14, 2015 at 0:14
  • $\begingroup$ @Moya : Minor confirmation: $L^+$ is just notation for the class of integrable (and thus $L^1$ since $f$ int. iff $|f|$ int.) non negative functions (hence the '+') right? $\endgroup$ Commented Mar 7, 2020 at 11:34

1 Answer 1

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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 $n\in \mathbb{N}$. So $\sup_{k\geq n}\int f_k \leq \int f$ for all $n\in \mathbb{N}$. Then it is clear 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
    Commented Nov 9, 2016 at 4:49
  • $\begingroup$ @Ninja Yes. Thanks for pointing it. I have corrected it. $\endgroup$
    – Ramiro
    Commented Nov 10, 2016 at 0:05
  • $\begingroup$ @Ramiro Where does the proof use the condition $f_j\leq f_{j+1}$? $\endgroup$
    – user469065
    Commented Oct 10, 2019 at 23:21
  • $\begingroup$ @user469065 : Late but: the condition says that $(f_n)$ i a monotone increasing sequence, and $f$ is its limit, so we know that $f_n\leq f$ for all $n$, and hence by monotonicity of $\int$, that $\int f_n\leq \int f$, which is used right before (1). $\endgroup$ Commented Mar 8, 2020 at 15:01
  • $\begingroup$ Could someone explain why it is clear that $\lim_{n \rightarrow \infty} \sup_{k \geq n} \int f_k \leq \int f$? Is it because since $f_n \leq f$ for every $n$ then it is certainly true for the supremum as well. Then if this is true for the supremum it is true on a smaller set ($k \geq n$) and finally, we take the limit and get the result. $\endgroup$ Commented Mar 29, 2021 at 22:41

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