show that $f=0$ a.e. given that $f$ is a bounded, measurable function on E and $\int_{E}{f}=0$ $f$ is nonnegative

Question

I am quite confused with this problem. This is somehow similar to a proposition following the Chebychev's Inequality. However, the latter doesn't require to $f$ to be bounded.

Do I just use similar approach but intead of having $E_n=\{x\in E:f(x)>=1/n\}$, I would have $E_n=\{x\in E: f(x)<=1/n\}$? Help, I am confused with the boundedness of f.

Should I use the Simple Approximation Lemma that assumes $f$ to be bounded? How do I go about with this?

The statement in the title is obviously false. Say $E=[0,2]$ (with Lebesgue measure) and set $f=1$ on $[0,1]$, $f=-1$ on $(1,2]$. — David C. Ullrich, Sep 30 '15 at 13:25
What's the problem? Assuming that $f=0$ except on $N$ and $f$ is bounded ($|f|<K$) you have $|f| < K\chi_N$ and $\int \chi_N = 0$. — skyking, Sep 30 '15 at 13:26
desperatemuch, are you perhaps missing a hypothesis that would make a true statement? — Sinister Cutlass, Sep 30 '15 at 13:34
This was not a leading question; it was an honest one. It's been a while since I did graduate analysis. I just wanted to make sure you had transcribed the problem completely to the Math Stack Exchange. — Sinister Cutlass, Sep 30 '15 at 13:37
I have no idea what you mean by "it should be integral not int". The integral of the example I gave is $0$, but the function is not $0$ almost everywhere. What are the actual hypotheses? What do you mean by "it should be integral not int"? — David C. Ullrich, Sep 30 '15 at 13:38
@SinisterCutlass oh I thought people mistook it as int = integer. That is all that there is, it says Let $f$ be a nonnegative bounded measurable function on a set E of finite measure. If $\int_E f = 0$. Show that f=0 a.e. on E — desperatemuch, Sep 30 '15 at 13:42
@desperatemuch dear, I think your question need $\color{red}{f\geq 0}$ too — R.N, Sep 30 '15 at 13:48
Yes, Razieh's suggestion would make the solution practically immediate. — Sinister Cutlass, Sep 30 '15 at 17:32

desperatemuch · Answer 1 · 2015-09-30T13:46:22.413

0

Since $f# is bounded then I could use the definition of the integral of bounded functions, that is, $\int_{E}{f} = inf{\int_{E} \psi : \psi simple, f<=\psi on E} $ which is equal to 0 since $\int_{E}{f}$ is 0. Then $\int_{E}{\psi} = 0$ Thus, $\psi = 0$ and therefore $f=0$ almost everywhere on E.

But I am not convinced with my own work...

edited Sep 30 '15 at 13:46

answered Sep 30 '15 at 13:41

desperatemuch

21

That's not the definition of the integral. The result you're trying to prove is false! I gave a simple counterexample. A really simple counterexample - what about the example don't you get? – David C. Ullrich Sep 30 '15 at 13:44
Please tell me if I get this right. The example that you gave implies that $\int_{[0,2]}{f} = \int_{[0,1]}{f} + \int_{(1,2]}{f} = 1+ (-1) = 0$ but $f$ cannot be negative 1 since f should be nonnegative. – desperatemuch Sep 30 '15 at 13:51
Oh, $f$ should be nonnegative! You didn't tell us that until now... that's different. – David C. Ullrich Sep 30 '15 at 13:53
sorry, i missed out on it. – desperatemuch Sep 30 '15 at 14:02

score 0 · Answer 2 · answered Sep 30 '15 at 13:50

0

The missing hypothesis in your statement is that $f$ must also be assumed to be nonnegative.

In this case, it is the contrapositive of the following statement:

If $f$ is a bounded measurable function such that $f>0$ on a set $E$ of positive measure, then $\int_E f>0$.

To prove this, let $A_n=\{x:f(x)\ge 1/n\}$. If there is some $A_n$ with positive measure, then $$\int_E f\ge\int_{A_n} f\ge\int_{A_n}1/n=\frac{\mu(A_n)}n>0,$$ otherwise the set $E=\{x:f(x)>0\}=\bigcup_{n\in\Bbb N}A_n$ is a countable union of nullsets and hence null, contradicting the assumption.

answered Sep 30 '15 at 13:50

Mario Carneiro

27,438

Two comments: (i) surely it's a much better idea, teaching a man to fish, etc, to get the OP to look up the question himself and determine what hypothesis he omitted from his post? (ii) I don't see how you can be so certain that's the missing hypothesis; I can think of another possibility... – David C. Ullrich Sep 30 '15 at 13:51
@DavidC.Ullrich Mainly because I've seen this problem before, and it's one of my favorite measure theory proofs. I don't know what name the theorem goes under, though, so I had some trouble finding duplicates, and I'm sure the OP might have the same issue. What is your alternate possibility? (cf. "teach a man to fish", my preferred response to a user who is having trouble with understanding the basics of the problem is "teach the hell out of it", with an answer so over-complete that it would be inappropriate if copied onto homework, and hopefully with enough context to clear up the confusion.) – Mario Carneiro Sep 30 '15 at 13:56
@Mario Carneiro I somehow came up with that but I am quite confused with the use of $A_n$ why should I be using $f>=1/n$? Can i not use $f<=1/n$?. f should be bounded, that means $|f|<M$ for some M>0 – desperatemuch Sep 30 '15 at 13:57
@MarioCarneiro My other possible hypothesis is obsolete now that he's clarified things. When it was still "int $f$ equals $0$ on $E$" it seemed possible that that was a garbled version of "$\int_F f=0$ for every $F\subset E$". – David C. Ullrich Sep 30 '15 at 13:59
I am sorry. This isn't homework. I am just so confused with the Lebesgue integral, I have this tendency to try to make out things based on the definition or theorems but I couldn't make this one out. There is a proposition somehow similar to this that is proved using Chebychev's inequality but it didn't specify f to be bounded. This problem here specifically requires $f$ to be bounded. Hence my confusion on what steps to take – desperatemuch Sep 30 '15 at 14:00
@desperatemuch Actually it doesn't matter that $f$ is bounded, although it does mean that no infinities show up in the inequalities, which simplifies matters. The reason we are using $f\ge1/n$ is because we want to capture the exceptional values where $f$ is far from zero, so that we can integrate over them to get a positive value. Also, if $f(x)>0$ then $f(x)\ge 1/n$ for some $n\in\Bbb N$, this is where the final decomposition of $E=\bigcup_nA_n$ comes from. – Mario Carneiro Sep 30 '15 at 14:02
@desperatemuch I'm not saying that this is homework, that was just an elucidation of my general strategy for this kind of problem, where the OP (you) is having as much difficulty with the concepts as with the problem itself. – Mario Carneiro Sep 30 '15 at 14:03
@MarioCarneiro yes and no, imo. Yes, and a big yes: The traditional "what have you done so far" is just silly when the OP is obviously totally lost. But no, regardless of how totally lost the OP is, it seems to me that we should at least require him to state the problem correctly, instead of telling him what his question actually was. Cuz he could go back and find the correct statement, and realizing that little details like what the hypotheses are can make a difference is a good thing to realize... – David C. Ullrich Sep 30 '15 at 14:07
the thing that is confusing me is since in class, my prof always asks, about the hypotheses. he often says, since $f$ is bounded, then what can we make out from there? – desperatemuch Sep 30 '15 at 14:09
1

Here's another way to see that the boundedness assumption is not necessary: suppose we've proven your theorem, assuming $f$ is bounded. Then for any (possibly unbounded) measurable function $g$ on $E$ such that $\int_Eg=0$, setting $f=\min(g,1)$ we have $0\le f\le g$ so $\int_Ef=0$ and the theorem says that $f=0$ a.e., hence $g=0$ a.e. (because if $min(g(x),1)=0$ then $g(x)=0$). Just because an assumption is present in a problem doesn't mean you have to use it. – Mario Carneiro Sep 30 '15 at 14:17

show that $f=0$ a.e. given that $f$ is a bounded, measurable function on E and $\int_{E}{f}=0$ $f$ is nonnegative

2 Answers2