I do not think this is true,

but at the same time I am not sure.

I know that if we assume that f is continuous instead of integrable then this statement is true. I just do not know how to provide a counterexample if it is false to show that this is wrong. Integrable does not imply continuity I know that much.

  • 1
    $\begingroup$ The simplest counter-example that you could possibly conceive of is $f\equiv 0$ except at possibly one point. This will indeed suffice. $\endgroup$ – Robert Wolfe Nov 26 '18 at 5:19
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    $\begingroup$ The intuition here that examples come from is that Lebesgue integration is “blind” to sets of measure zero: if $f$ is integrable over $E$, and $Z\subseteq E$ has zero measure, then $\int_E f = \int_{E-Z} f$. Getting “everywhere” information from a tool that can only give you “almost everywhere” information is highly unlikely. $\endgroup$ – Santana Afton Nov 26 '18 at 5:30

Here is a counterexample using Lebesgue integration.

Let $f=\chi_{\Bbb Q\cap [a,b]}$. Then $f$ is $0$ a.e. , hence Lebesgue integrable. Next for any Lebesgue integrable $g$ we have $fg=0$ a.e. so that $fg$ is Lebesgue integrable and $\int_a^b fg=0$ . But $f$ is not identically zero in $[a,b]$.


The conclusion should by $f(x)=0$ almost everywhere. To prove, set $g$ equals the sign of $f$ times $f$, which is integrable and $fg=|f|$. Then you get that $\int|f(x)|dx=0$, which implies that $f=0$ almost everywhere.

  • $\begingroup$ The sign of $f$ times $f$ is just $|f|$, isn't it? So $fg=|f|f$. $\endgroup$ – TonyK Nov 28 '18 at 11:15

Okay since this is true for every integrable function g, take $g=f$

Now this becomes $\int^{b}_aff=0$

Now we know that $f^2(x)\geq0$

and $\int^{b}_af^2(x)=0$, This means that the area of the curve is zero even if the function is always above x axis, only one conclusion can be derived from this, that is

$f^2(x) \equiv 0$ $\forall$ x $\in$ $(a,b)$

The function becomes identically zero.

therefore $f(x)=0,\forall x\in[a,b]$

EDIT: If it is given that f(x) and g(x) is continuous functions then this approach would work, since there is no such condition in this question, hence the statement becomes false.

  • $\begingroup$ whoever down voted my answer, could you please explain what is wrong in this?? why did i get downvoted? $\endgroup$ – Rohit Bharadwaj Nov 26 '18 at 17:03
  • $\begingroup$ You should read the given counter examples and figure out that the statement is not correct and must be modified. How would you prove a false statement?! $\endgroup$ – M. Rahmat Nov 26 '18 at 21:18
  • $\begingroup$ @M.Rahmat but the author said he's not sure whether this is true or not, and i don't think there is anything wrong in my approach to prove this, this same question was done in my class few days back with the same approach and i haven't been taught about Lebesgue integration yet so i can't understand the accepted answer. $\endgroup$ – Rohit Bharadwaj Nov 27 '18 at 14:52
  • $\begingroup$ Bharadwaj. Define a function $f$ that is 1 at x=-1, 0 and 1 and is 0 every where else. The integral of this function from -2 to 2 is 0, but the function is not 0. It is 0 almost everwhere. Edit your answer I will upgrade your answer. $\endgroup$ – M. Rahmat Nov 27 '18 at 21:46
  • $\begingroup$ Thank you @M.Rahmat, I realized that it is false, but i wonder why they solved it like this in our class, anyways thank you for the example, $g(x)=1$ and $f(x)$ as defined by you will not make the statement true $\endgroup$ – Rohit Bharadwaj Nov 28 '18 at 5:19

Another counter example is that consider $f:[0,1]\rightarrow [0,1]$ given by $f(x)=0$ if $x$ is irrational or $x=0$ and $f(\frac{p}{q})=\frac{1}{q}$ where $p\in \Bbb Z-\{0\},q\in \Bbb N,gcd(p,q)=1$ , then $f$ is Riemann integrable and $\int_0^1 f=0$ and for any other Riemann integrable $g$ we have using Cauchy-Schwarz inequality $$|\int_0^1 fg|^2\leq\int_0^1 f^2 ×\int_0^1 g^2\leq \int _0^1 f×\int_0^1 g^2=0×\int_0^1 g^2=0$$ ,hence $\int_0^1 fg=0$ . But $f$ is not identically zero in $[0,1]$.

Though I used $[0,1]$ as a special interval , by slide modifications you can give argument for general compact interval.Note one thing is that I have only consider Riemann integration i.e. here is no Lebesgue integration. You can prove using only definition of Riemann integration that $\int_0^1 f=0$


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