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The following question is taken from Royden Real Analysis $4$th edition, Chapter $4,$ question $20.$

Let $\{f_n\}$ be a sequence of nonnegative measurable functions that converges to $f$ pointwise on $E.$ Let $M\geq 0$ be such that $\int_E f_n\leq M$ for all $n.$ Show that $\int_E f\leq M.$ Verify that this property is equivalent to the statement of Fatou's Lemma.

I have proven that $\int_E f\leq M.$ But I have no idea on how to tackle second part.

After some googling, I found a solution in MSE, which goes as follows:

Let $(f_n,n\in\Bbb N)$ be a sequence of measurable integrable functions and $a_N:=\inf_{k\geqslant N}\int f_kd\mu$. Working with the sequence $(f_n,n\geqslant N)$ (for which the sequence of integrals has the same $\liminf$ as those of the whole sequence), one can see that that $\int fd\mu\leqslant a_N$ for each $N$. Now take the limit $\lim_{N\to +\infty}$.

Questions:

$(1)$ What is a motivation of considering $a_N:=\inf_{k\geq N}\int f_k d\mu?$

$(2)$ Why does $\int f d\mu\leq a_N$ for each $N?$

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    $\begingroup$ The motivation for $a_N$ is that it is part of the definition of $\liminf$. That $\int f \ d\mu \leq a_N$ follows from the fact that the subsequence $(f_k)_{k \geq N}$ is itself a sequence, so that the conclusion follows from the hypothesis. $\endgroup$ – Project Book Dec 28 '17 at 3:28
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$\displaystyle\int\inf_{k\geq N}f_{k}d\mu\leq\int f_{k}d\mu$ for all $k\geq N$, so $\displaystyle\int\inf_{k\geq N}f_{k}d\mu\leq a_{N}\leq M:=\sup_{N}a_{N}$ for all $N=1,2,...$. Now $\inf_{k\geq N}f_{k}\rightarrow\liminf_{n}f_{n}$ as $N\rightarrow\infty$, by the result, then $\displaystyle\int\liminf_{n}f_{n}d\mu\leq M$, and it is actually the case that $M=\sup_{N}a_{N}=\sup_{N}\inf_{k\geq N}\displaystyle\int f_{k}d\mu=\liminf_{n}\displaystyle\int f_{n}d\mu$.

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  • $\begingroup$ The reason for that... I would say, because it is part of the form of Fatou's Lemma... $\endgroup$ – user284331 Dec 28 '17 at 3:22
  • $\begingroup$ So I think your number 2) does not necessarily hold, but I switch to another strategy. $\endgroup$ – user284331 Dec 28 '17 at 3:26

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