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Assume $f$ is Riemann integrable and further assume that $\int_a^x f=0$ for all $x$. How would I go about showing that $f$ itself is $0$ almost everywhere?

I am new to Lebesgue's measure theory so I am hoping for a somewhat elementary proof if possible?

I know that almost everywhere means all except a set of measure zero. I was wondering if I could get a point to start on? We are told $f$ is Riemann integrable, so that means by Lebesgue's criterion for Riemann integration that there are at most countably infinitely many discontinuities. Thank you!

Tomas
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We have $\int_x^y f = \int_a^y f - \int_a^x f = 0$. Now assume $f(b) > 0$ for some $b$ and that $f$ is continuous at $b$ (otherwise we're still in a set with measure $0$ by Lebesgue's criterion). Then there is $\varepsilon > 0$ with $f(y) > f(b)/2$ for all $y \in (b - 2\varepsilon, b + 2\varepsilon)$. In patricular, since $[b - \varepsilon, b + \varepsilon] \subseteq (b - 2\varepsilon, b + 2\varepsilon)$, we have $f([b - \varepsilon, b + \varepsilon]) \subseteq [f(b)/2, \infty)$. Thus $$\int_{b - \varepsilon}^{b + \varepsilon} f \geq \int_{b - \varepsilon}^{b + \varepsilon} f(b)/2 = 2 \varepsilon f(b)/2 = \varepsilon f(b)> 0.$$ Contradiction.

  • Hi, thanks for your answer. May I ask how you came to the conclusion that there is $\epsilon>0$ with $f(y)>f(b)/2$ for all $y$ in that epsilon ball? And how you know that the image of the epsilon ball under $f$ is a subset of that interval? – Tomas Jun 24 '16 at 14:34
  • Also, actually, why is this last statement a contradiction? In our initial assumption, we assumed that $f(b)>0$, so why is $\epsilon f(b)>0$ not possible? – Tomas Jun 24 '16 at 16:26
  • Existence of $\varepsilon$ is the continuity of $f$ at $b$. The contradiction is $0 = \int_{b - \varepsilon}^{b + \varepsilon} f > 0$. And I fixed a small mistake. –  Jun 25 '16 at 08:43
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    Assuming continuity hugely restricts the setting of the question. – Did Jun 25 '16 at 08:50
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    The set at which $f$ is not continuous is a set with measure zero. Thus it's sufficient to just show $f(x) = 0$ for $f$ being continuous at $x$. That's what I do. –  Jun 25 '16 at 08:53
  • @Did, why is that? Function $f$ is assumed to be Riemann integrable, which implies that $f$ is continuous a.e. – Glinka Jun 25 '16 at 13:01
  • @Glinka Why is what? Almost everywhere continuity being the heart of the matter should be mentioned explicitely. – Did Jun 25 '16 at 13:45
  • @Did, indeed it should be. It's called the Lebesgue criterion for Riemann integrability, quite common. But there's no restrictions of the setting of the question, as you wrote; the answerer just forgot to specify the criterion he used (or considered it obvious and not worthy to mention, which wasn't right of him). – Glinka Jun 25 '16 at 19:28
  • He mentioned Lebesgue's criterion in his question. Therefore I assumed that it's understandable. –  Jun 25 '16 at 19:29
  • Oh, you're right, sorry... I'm being unobservant. Still wouldn't hurt to mention for possible future readers and for OP convenience (considering it raises questions even from experienced users like @Did) – Glinka Jun 25 '16 at 19:35
  • I'm rusty in my Real Analysis for now, but isn't it much more simple to just say "suppose f is not 0 almost everywhere. Let's assume x > a. Then there is a subset A of (a, infinity) with positive measure for which f(x) does not equal zero for all x in A. Thus, for any x in the interior of A, we have that the integral of f from a to x is not equal to 0, ending the proof via contraposition." ? – change_picture Jun 25 '16 at 19:57
  • @Glinka I already added it. –  Jun 25 '16 at 20:19
  • @gorzardfu Could you explain your steps more precisely: i.e. why does the set A have positive measure. Why does it have non-empty interior. (I know that it's true, but that's where my arguments are actually needed) –  Jun 25 '16 at 20:20
  • @menag Because what it means for a function to be 0 almost everywhere is that the set over which f is not zero has measure 0. If f was not 0 almost everywhere, then there is a set A of positive measure over which f is not 0. – change_picture Jun 25 '16 at 20:36
  • @gorzardfu, interior of $A$ can be empty, for example if $A=[a,b]\setminus\mathbb{Q}$, for any segment $[a,b]\subset\mathbb{R}$, $a\neq b$ – Glinka Jun 25 '16 at 21:53
  • Good point @Glinka. – change_picture Jun 25 '16 at 22:06
  • And I'm sorry @menag I didn't read your entire question. But I believe there is a way to fix my proof, yes? I'm just saying the existence of such a positive measure set A with all values, lets say, greater than a, would have a subset B with positive yet finite measure. Let x = sup(B). Then integrating f from a to x gives you something nonzero. That's my best attempt for fixing my proof. – change_picture Jun 25 '16 at 22:12
  • There are still a few issues. The values between $a$ and $x$ could still be positive and negative and integrating over them could result in zero. Further, the supremum does not have to exist - $B$ has finite (positive) measure but doesn't have to be bounded. You should really use the continuity of $f$ almost everywhere. –  Jun 26 '16 at 07:44
  • Posted an answer that doesn't use Riemann integrability, check it out. @gorzardfu, you can consider it a fix for your proof. – Glinka Jul 05 '16 at 13:42
  • Yes @menag you're right. My proof certainly needed revision but it wasn't really a proof but more of a (poorly) drafted outline of a proof. – change_picture Jul 05 '16 at 21:01
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The statement remains true without Riemann integrability constraint — Lebesgue integrability of $f$ is sufficient (as was suggested by gorzardfu). So let $\int_a^xf=0$ for all $x\in\mathbb{R}$. This implies that $\int_y^zf=0$ for any $y,z\in\mathbb{R},\ y<z$, as $\int_y^zf=\int_a^zf-\int_a^yf$, as was noted in menag's answer. So $\int_If=0$ for any interval $I\subset\mathbb{R}$, and so for any Borel set, and hence for any measurable set.

Suppose $f>0$ on some set $A\subset\mathbb{R}$ of strictly positive measure. That means that for some $\varepsilon>0$ we have $f>\varepsilon$ on some $B\subset A$ of strictly positive measure and so $\int_B f>\mu(B)\cdot\varepsilon>0$, which is a contradition.

Another way of looking at this is to define $g(x)=\int_a^xf$. Then $f$ is a weak derivative of $g$. On $\mathbb{R}$ it implies, in particular, that $g$ is differentiable almost everywhere and $g'=f$ a.e.. And since $g\equiv0$ on $\mathbb{R}$, we have $f=0$ a.e..

Glinka
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