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Definition of Lebesgue measurable function: Given a function $f: D \to \mathbb R \cup \{+\infty, -\infty\}$, defined on some domain $D \subset \mathbb{R}^n$, we say that $f$ is Lebesgue measurable if $D$ is measurable and if, for each $a\in[-\infty, +\infty]$, the set $\{x\in D \mid f(x) > a\}$ is measurable.

If $f$ is an extended real valued(codomain is $[-\infty, +\infty]$) measurable function defined on $[a, +\infty)$ and it's Lebesgue integrable with $\int_{G} f(x) dx \ge 0$ for $\forall$ open set $G \subset (a, +\infty)$, how about its integral on a $G_{\delta}$(a countable intersection of open sets) set? Is still $\int_{G_{\delta}} f(x) dx \ge 0$?

Besides, if domain of $f$ is modified to $\mathbb R$ that is $f$ is a real valued measurable function defined on $\mathbb R$ with the same property, will the conclusion still be true?

Eugene Zhang
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Bear and bunny
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  • What if your $G_\delta$ set has the form $\bigcap U_n$ where $U_n$ are open and $U_1 \supseteq U_2 \supseteq U_3 \supseteq \dots$ – GEdgar Sep 06 '15 at 22:10
  • @GEdgar: Hmmm, I'm not sure. $\bigcap U_n$ can be closed or open or neither with respect to your assumption? – Bear and bunny Sep 06 '15 at 22:15
  • There is some "monotone convergence theorem" right? – GEdgar Sep 06 '15 at 22:17
  • @GEdgar: Yes. So $\int_{U_i} \ge \int_{U_{i+1}}$. How to do next? – Bear and bunny Sep 06 '15 at 22:20
  • I'm guessing this problem is too advanced for you. Maybe consult the instructor for help. – GEdgar Sep 06 '15 at 22:22
  • @GEdgar: Hmmm. Thanks. – Bear and bunny Sep 06 '15 at 22:24
  • @GEdgar: Without assuming $f \geq 0$, the monotone convergence theorem is not applicable. But since we are assuming that $f$ is integrable (at least it seems that's what the OP is assuming), we can use the dominated convergence theorem. Also note that each Lebesgue measurable set is (up to a set of measure zero) a $G_\delta$ (at least if it is of finite measure) so actually we can show that $f \geq 0$ almost everywhere. – PhoemueX Sep 07 '15 at 05:48
  • @PhoemueX: What is OP? – Bear and bunny Sep 07 '15 at 13:41
  • @Bearandbunny: I am using it (and have seen it used) for "original poster" or sometimes "original post". – PhoemueX Sep 07 '15 at 15:06
  • @PhoemueX: Yes. Actually I was asked to show $f \geq 0, a.e.$ in the exercise. I know how to do it and $\int_{G_{\delta}} f(x) dx \ge 0$ is the key step about which nevertheless I've no idea. So I post here for help. – Bear and bunny Sep 07 '15 at 15:15
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    Be careful with your title! $f(x) = x$ does not have a strictly positive integral on every $G_\delta$ set. – Carl Mummert Dec 09 '15 at 23:46
  • @CarlMummert: Because $G_\delta$ can have only one element? – Bear and bunny Dec 10 '15 at 22:30

3 Answers3

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The answer is yes.

Let $U_n\:(\:n\in\Bbb{N})$ be countable number of open sets. Then any $G$-set $$ G_{\delta}=\bigcap_{n=1}^{\infty}U_n=\bigcap_{n=1}^{\infty}\bigcap_{k=1}^{n}U_k=\bigcap_{n=1}^{\infty}O_n $$ where $O_n=\bigcap_{k=1}^{n}U_k$. Clearly any $O_n$ is open set and $$ O_1\supset O_2\supset \cdots\supset O_n\supset \cdots $$ Since $O_n$ is open $$ \int_{O_n}f\:dx\geqslant0 $$ So $$ \int_{G_\delta}f\:dx=\int_{\bigcap_{n=1}^{\infty}O_n}f\:dx=\lim_{n\to\infty}\int_{O_n}f\:dx\geqslant0 $$ Here we use Monotone Class theorem in measure theory that $O_1\supset O_2\supset \cdots\supset O_n\supset \cdots$ implies $$ \mu \left ( \bigcap _{n=1}^{\infty} O_n\right )=\lim _{n\to \infty }\mu (O_n) $$ Since given $\epsilon>0$, there exists $M$ that for all $r>M$ $$ \left|\int_{\mathbb{R}} f\:dx-\int_{[-r,r]} f\:dx\right|<\epsilon $$ It holds for domain of $f$ in $\mathbb{R}$.

Eugene Zhang
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  • $\int_{O_n}f:dx\geqslant0$ for any $n \in \mathbb N^+$, then $\lim_{n\to\infty}\int_{O_n}f:dx\geqslant0$. How? coz $\int_{O_n}f:dx$ is non-increasing. – Bear and bunny Sep 07 '15 at 15:12
  • Let $∫_{O_n}fdx=\mu(O_n)$ and use the fact that if each $a_n>0$, then $\lim a_n\ge 0$. – Eugene Zhang Sep 07 '15 at 16:15
  • Ok. But actually $a_n \ge 0$, it doesn't matter? Otherwise, I really want to upvote your idea when such a piece of nuisance gets removed. – Bear and bunny Sep 07 '15 at 19:22
  • It should be $a_n\ge0$ and is a typo. I'd be appreciated if you accept the answer. – Eugene Zhang Sep 07 '15 at 20:31
  • I leave a comment here so that help every new comer visiting this post. Hermes' answer is a directed solution to the problem and PhoemueX emphasizes the integrability is an important during proving. So combine them two together. – Bear and bunny Sep 08 '15 at 20:46
  • @Bear and bunny, it should be "is a direct solution", not "is a directed solution". – Eugene Zhang Sep 08 '15 at 21:47
  • ohhh, so sorry, my typo. reading too much papers talks about directed information recently. o(╯□╰)o. – Bear and bunny Sep 08 '15 at 22:08
  • There must be some finite measure space hypothesis here, otherwise $O_n=(n,\infty)$ provides a counterexample. – Ian Dec 09 '15 at 23:41
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Let $G = \bigcap_n U_n$ be a $G_\delta$. As noted already by others, by switching to $V_n =\bigcap_{i=1}^n U_n$ (this is still open with the same intersection of all sets), we can assume $U_{n+1} \subset U_n$.

Now set $f_n := f \cdot 1_{U_n}$, where $1_A$ is the indicator function of $A$. I leave it to you to verify that $f_n \to f \cdot 1_G$ pointwise. Furthermore, we have $|f_n|\leq |f|$, where $|f|$ is integrable, since you assume that $f$ is integrable (this is an important point, see below).

Thus, the dominated convergence theorem yields

$$ \int_G f\,dx =\lim_n \int_{U_n} f \, dx \geq 0. $$

Now let us see that integrability of $f$ cannot simply be omitted. To this end, let $(x_n)_n$ be an enumeration of the rational numbers and set $$ f = \bigg( \sum_n \frac{1}{|x-x_n|}\cdot 1_{B_{2^{-n}}(x_n)}(x) \bigg) - 1_{\overline{B_N (0)} \setminus \bigcup_n B_{2^{-n}}(x_n)} $$ for a suitable (large) $N \in \Bbb{N}$. It is then not hard to verify $$ \int_U f \, dx =\infty \geq 0 $$ for every nonempty open set $U$. Nevertheless, we do not have $f \geq 0$.

Even more, for the closed (hence $G_\delta$) set $$ A := \overline{B_N (0)} \setminus \bigcup_n B_{2^{-n}}(x_n), $$ we have $\int_A f \, dx <0$.

In this example, $B_r (x)$ is the open ball of radius $r$ around $x$.

PhoemueX
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  • I'm not sure what does this mean $f = \bigg( \sum_n \frac{1}{|x-x_n|}\cdot 1_{B_{2^{-n}}(x_n)}(x) \bigg) - 1_{\overline{B_N (0)} \setminus \bigcup_n B_{2^{-n}}(x_n)}$. Why do you use this? – Bear and bunny Sep 07 '15 at 19:23
  • I've noticed $f_n := f \cdot 1_{U_n}$ that you leaved to me to verify its pointwise convergence. Is the $f$ arbitrary? – Bear and bunny Sep 07 '15 at 19:28
  • Yes, $f$ could be arbitrary but it actually is your $f$, ie the one which satisfies $\int_U f , dx \geq 0$ for every open $U$. I somehow neglected that you assume $f$ to be defined on an interval, but that does not change much. You can ignore the special definition of $f$ later on ($f =\sum_n \dots$) if you want. This is just to show that the claim is false in general if we do not assume $f$ to be integrable. – PhoemueX Sep 08 '15 at 01:51
  • Let me first answer the verification, pointwise convergence $f_n \to f$. I try to use definition of pointwise convergence that is $\forall x \in \mathbb R, \forall \epsilon > 0, \exists N \ge 1, s.t, |f_n - f \dot 1_{G}| < \epsilon, \forall n \ge N$. Then when $x \in G, |f_n - f 1_{G}| = 0, \forall n \in \mathbb N^+$; when $x \not\in G$, due to $x \not\in \bigcap_{n} U_n$, does $\exists N \in \mathbb N^+, s.t, |f_n - f \dot 1_{G}| = 0$, when $n \ge N$ and N should depend on $U_n$ which $x$ stays(that is depends on $x$). Done. – Bear and bunny Sep 08 '15 at 02:23
  • $f$ integrable is presumed. Why need to mention it? – Eugene Zhang Sep 08 '15 at 05:08
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    @hermes: 1) Because I was interested on whether the theorem still holds without integrability. 2) Because it shows that any answer which does not use the integrability at some point is wrong. – PhoemueX Sep 08 '15 at 05:21
  • Any answer must use the integrability for $\int_{O_n}f:dx\geqslant0$ is used. – Eugene Zhang Sep 08 '15 at 06:14
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    @hermes: We could also have $\int_{O_n}f , dx = \infty$ if we do not assume that $f$ is integrable (only quasi-integrable, i.e. $\int f_{-} , dx < \infty$). – PhoemueX Sep 08 '15 at 08:19
  • What is $\overline{B_N (0)}$ ? Closure of $B_N (0)$? – Bear and bunny Sep 08 '15 at 13:24
  • Besides, $B_{N(0)}$ means ball centered at 0 with radius N? – Bear and bunny Sep 08 '15 at 13:33
  • @Bearandbunny: yes, exactly. – PhoemueX Sep 08 '15 at 14:48
  • I don't know why $\int_U f , dx =\infty$. The function is a little bit complicated, can you explain a little bit? I appreciate. – Bear and bunny Sep 08 '15 at 14:59
  • I think $ \overline{B_N (0)} \setminus \bigcup_n B_{2^{-n}}(x_n)$ = empty coz rationals are dense in $B_{N(0)}$. Then $\bigcup_n B_{2^{-n}}(x_n)$ will actually cover $\overline {B_{N (0)}}$. – Bear and bunny Sep 08 '15 at 15:07
  • @Bearandbunny: So, the set is not empty because (this is a beautiful argument) $\lambda(\bigcup_n B_{2^{-n}}(x_n)\leq \sum_n \lambda(B_{2^{-n}}(x_n) <\infty$, so that (at least if we take $N$ large enough), the difference set is is positive measure. To see $\int_U f ,dx =\infty$, note that each open set contains a full interval $(x_n -\epsilon, x_n +\epsilon)$ for some $n$ and some $\epsilon >0$ (since rationals are dense). Furthermore, $\int_I f \dx \geq \int_I 1/|x-x_n|, dx =\infty$ for any open interval which contains $x_n$. – PhoemueX Sep 08 '15 at 16:56
  • Awesome. I agree with. But why did you set a minus part in $f$ that is $1_{\overline{B_N (0)} \setminus \bigcup_n B_{2^{-n}}(x_n)}$. coz I think any interval contains some elements of $\overline{B_N (0)} \setminus \bigcup_n B_{2^{-n}}(x_n)$ and since it will be substract from $f$ when doing integration(I'm not sure whether the subtract part will have an integral of $\infty$ on any interval $I$), will this part affect the integral result of $\infty$? – Bear and bunny Sep 08 '15 at 17:48
  • @Bearandbunny: I intersected with $B_N (0)$ so that this does not happen. The ball has finite measure, so that the subtracted part has finite integral. If we only integrate over some subset (interval), the integral of course remains finite. Now $\infty -y=\infty$ for any finite $y$, so we have no problem. – PhoemueX Sep 09 '15 at 03:39
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The hypotheses imply $f\in L^1[a,\infty).$ By the Lebesgue differention theorem,

$$ f(x) = \lim_{h\to 0} \, \frac{1}{h}\int_x^{x+h} f$$

for a.e. $x \in [a,\infty).$ Since each integral on the right is nonnegative, so is its limit when it exists. Thus $f\ge 0$ a.e. in $[a,\infty).$ It follows that $\int_E f \ge 0$ for all measurable subsets $E\subset [a,\infty),$ never mind the concern over $G_\delta$ subsets.

zhw.
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