# Let $f$ be integrable on $[a,b]$ and suppose for each integrable function $g$ defined on $[a,b]$, $\int^{b}_afg=0$, then $f(x)=0,\forall x\in[a,b]$

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.

• The simplest counter-example that you could possibly conceive of is $f\equiv 0$ except at possibly one point. This will indeed suffice. – Robert Wolfe Nov 26 '18 at 5:19
• 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. – 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.

• The sign of $f$ times $f$ is just $|f|$, isn't it? So $fg=|f|f$. – 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.

• whoever down voted my answer, could you please explain what is wrong in this?? why did i get downvoted? – Rohit Bharadwaj Nov 26 '18 at 17:03
• 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?! – M. Rahmat Nov 26 '18 at 21:18
• @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. – Rohit Bharadwaj Nov 27 '18 at 14:52
• 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. – M. Rahmat Nov 27 '18 at 21:46
• 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 – 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 $$\int_0^1 f=0$$