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Suppose I have two Lebesgue measurable functions $f,g : [0,1] \to \mathbb R_+$ with the following property: $$ \int_0^x f(y) dy = \int_0^x g(y) dy$$ for a.e. $x \in [0,1]$. Can I conclude that $f=g$ a.e. on $[0,1]$ from here?

More generally, if I have two random variables $x,y$ such that $\mathbb E[ x \vert x \le a] = \mathbb E[y \vert y \le a]$ for a.e. $a \in \mathbb R$, can we say that $x =y$ a.s.?

avk255
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  • This is a nice corollary of the fact that if the integral of a function is zero a.e. then the function is identically zero a.e. Just consider the integral of the difference of the two functions. – rnrstopstraffic Dec 07 '19 at 22:01

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Let $f(x)=\frac 1 x$ for $x >0$, $f(0)=0$ and $g=2f$. Then hypothesis is satisfied but $f \neq g$. (Both sides are $\infty$.)

However the conclusion is true if $f$ and $g$ are integrable. This is a consequence of Lebsgue's Theorem. Ref.: https://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem

Kavi Rama Murthy
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  • If g=2f, I don't see how their integral can be equal, plus their integral don't even exist if f(x)=1/x – Jeanba Dec 02 '19 at 08:55
  • In measure theory one is not afraid of $\infty$!. Both sides of the equation exist and the value of either side is $\infty$. @Jeanba – Kavi Rama Murthy Dec 02 '19 at 08:58
  • Ok thanks for the precision! – Jeanba Dec 02 '19 at 09:52
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    For the case that $f,g$ are integrable, one can also give a proof without using the Lebesgue differentation theorem. See https://math.stackexchange.com/questions/16244/if-int-0x-f-dm-is-zero-everywhere-then-f-is-zero-almost-everywhere/3458702#3458702 – PhoemueX Dec 02 '19 at 20:18
  • Thanks everyone. Indeed, I missed integrability in the condition. The proof without the Lebesgue differentiation theorem is quite nice too. – avk255 Dec 03 '19 at 10:43