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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!


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

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..


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.

  • $\begingroup$ 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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