$f(x)$ is positive and continuous function on $\mathbb{R}$ and, moreover, $\int_{-\infty}^{+\infty}f(x)dx=1$. $\alpha\in(0;1)$ and $[a;b]$ is the interval having a minimum length such that the $\int_{a}^{b}f(x)dx=\alpha$. Prove $f(a)=f(b)$.

With mean value theorem easy to show, that for every $\alpha\in(0;1)$, exist such $[a;b]$, so $\int_{a}^{b}f(x)dx=\alpha$. The statement looks like Rolle's theorem, but I have no idea. I will be grateful for help.


Suppose $\int_a^b\,f(x)\,\text{d} x = \Phi(b)-\Phi(a)$.

Then, suppose we have that interval $[a_0,b_0]: \int_{a_0}^{b_0}\,f(x)\,\text{d} x =\alpha$, let it be $b_0-a_0=c$.

Let $g(a)=\int_a^{a+c}\,f(x)\,\text{d} x$. We have $g(a_0)=\alpha, g(a)<\alpha, a\neq a_0$.

$g'(a)=\lim_{\Delta x \to 0}\dfrac{\Phi(a+c+\Delta x)-\Phi(a+\Delta x)-\Phi(a+c)+\Phi(a)}{\Delta x}=f(a+c)-f(a)$

So it is differentiable and it has maximum at $a_0$, so the derivative is equal to zero.


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