# Integral equality when $f(x) \ge g(x)$

Given that $f(x) \ge g(x)$ for $x$ in $[0,1]$, I need to find an example of two different functions such that $$\int_{0}^1 f(x)\,dx = \int_{0}^1 g(x)\,dx.$$

Edit: my answer was to take a function, like f(x) = 1 and g(x) will be the same with discontinuous one point at, say, x=$\frac{1}{2}$. but it seems too obvious for an exam answer...

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How do you want the integral inequality to be ? – Amr Jun 15 '13 at 14:43
Welcome to Math.SE. Please, give us two further pieces of information: 1) the context of the problem and 2) what you have already tried / where you are getting stuck. – Nick Peterson Jun 15 '13 at 14:44
discontinuous functions – eccstartup Jun 15 '13 at 14:46

Hint This is not possible for two continuous functions (why?). So at least one of the functions needs to be discontinuous at at least one point.

Hint 2 For the functions to be different, they don't need to be different at many points.

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The two conditions together imply that $f$ and $g$ can only differ on a null-set. So at least one of them will not be continuous. One way to achieve what you want is to take any continuous function $g$ and redefine it at one point to obtain $f$.

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This was my answer, but i want to find an example that has "more" difference between the functions – john_gayl Jun 15 '13 at 14:50
Then redefine the function on countably many points. But you have to be careful on how you do it, because it might end up not being Riemann integrable. If you redefine on a finite number of points, then you are good for sure. – Martin Argerami Jun 15 '13 at 14:51
@רתםברגר Perhaps the Cantor set (which is uncountable and of measure zero) is large enough for your liking? – Lord_Farin Jun 15 '13 at 14:53

This would mean that $f=g$ almost surely but not everywhere.

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$f(x)= \begin{cases}0&\text{ if$x\in Q^c$} \\ \frac{1}{q} &\text{ if$x=\frac{p}{q}$and$p,q\in N$,gcd(p,q)=1}\end{cases}$

And $g(x)=0\forall x\in[0,1]$

Here $g$ are integrable and it can be shown that $f$ is also integrable and $\int_0^1f=\int_0^1g=0$

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