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This question already has an answer here:

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!

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marked as duplicate by Claude Leibovici, egreg, user491874, Guy Fsone, Did measure-theory Jan 27 '18 at 12:07

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  • $\begingroup$ Not countably many, a set of measure 0 (which can be uncountable). $\endgroup$ – zhw. Jul 6 '16 at 21:19
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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..

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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.

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  • $\begingroup$ 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? $\endgroup$ – Tomas Jun 24 '16 at 14:34
  • $\begingroup$ 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? $\endgroup$ – Tomas Jun 24 '16 at 16:26
  • $\begingroup$ 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. $\endgroup$ – Paul K Jun 25 '16 at 8:43
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    $\begingroup$ Assuming continuity hugely restricts the setting of the question. $\endgroup$ – Did Jun 25 '16 at 8:50
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    $\begingroup$ 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. $\endgroup$ – Paul K Jun 25 '16 at 8:53

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