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Let $f,g$ be Riemann integrable on $[0,1]$ such that $\int_0^1 f=\int_0^1 g=1$. Show that there exists $0\leq a<b\leq 1$ such that $\int_a^b f=\int_a^b g=\frac{1}{2}$.

If there is only one function $f$, it is easy. What about two functions?

xldd
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2 Answers2

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Denote by $u(a)$ the (smallest) number in $(0,1)$ such that $\int_a^{u(a)} f=1/2$. Use $v(a) $ for the analogous function with $g$. We can clearly define these for $a=0$. We then extend these to functions $u:[0,b] \to \mathbb{R}$ and $v:[0,c] \to \mathbb{R}$ where $b=u(0)$ and $c=v(0)$.

Note that $u, v$ are increasing.

If $u(0)=v(0)$ we are done. Suppose not.

Suppose WLOG that $u(0)<v(0)$. We have $\int_{u(0)}^{1} f=1/2=\int_{v(0)}^1g$. Therefore $u(b)=v(c)=1$. Since $b<c$ and $u, v$ are increasing, $u(b) \geq v(b) $.

By the intermidate value theorem, there is a point $a\in (0,b)$ at which $u(a) = v(a) $. By definition of $u, v$ we are done.

MathTrain
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    Ah, but then $u(a)$ need not be unique, since more than one point $p > a$ could satisfy the given property $\int_a^p f = \frac 12$. Would you choosethe infimum over such $p$? – Sarvesh Ravichandran Iyer Aug 27 '19 at 09:15
  • Sounds like a great solution! I'll correct the solution to reflect this – MathTrain Aug 27 '19 at 09:16
  • Yes, I think things are fine now. – Sarvesh Ravichandran Iyer Aug 27 '19 at 09:18
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    I think this is not quite complete yet. In order to apply the intermediate value theorem you need $u$ and $v$ to be continuous. If you pick functions $f$ and $g$ that are constant zero on some subintervals your definition of $u$ and $v$ doesn't always give a continous function (see the duplicate question for an example). – quarague Aug 27 '19 at 09:47
  • You are very right. I'm wondering if the problem could be avoided by either a) picking the supremum instead of the infimum or b) somehow editing the function to plug discontinuities with linear connections. I'll have to think about this more. Though I'm sure the duplicate (which I've flagged) has an answer, I'm still interested in coming up with my own – MathTrain Aug 27 '19 at 09:55
  • @MathTrain Why $u, v$ are increasing. If $f,g$ both are nonnegative, it is OK. But what about when it is not the case. – xldd Aug 29 '19 at 05:45
  • @xldd Since it is defined as an infimal value, they do not to be nonnegative for it to work. If the function $f$ goes negative, $u$ will simply jump forward to after it has become positive again and the integral has "made up" the lost area. This does pose a problem for continuity, but that's a problem we already had! – MathTrain Aug 29 '19 at 05:55
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This is a partial solution assuming that $f, g > 0$. There is probably some clever trick to generalize it but I don't see it right now.

Let $s \in (0,1)$ be the unique point where $\int_0^s f(x)dx=\frac{1}{2}$ and similarly $t \in (0,1)$ where $\int_0^t g(x)dx=\frac{1}{2}$. Consider the function $a \mapsto b_f(a)$ such that $\int_a^{b_f(a)}f(x)dx=\frac{1}{2}$ for all $a \in [0,s]$. Then $b_f(0)=s$, $b_f(s)=1$ and $b_f$ is continuous and strictly increasing. Define the function $b_g$ the same way for $g$ and $t$. These two functions intersect at some point and this gives you the desired $a$ and $b$.

quarague
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