$\int_0^1f=\int_0^1g=1$ implies $\int_If=\int_Ig=\frac{1}{2}$ on some interval $I\subset[0,1]$

Asked Oct 31 '23 at 15:14

Active Oct 31 '23 at 15:36

Viewed 75 times

I came across such a problem in my exercise:

Suppose $f$, $g$ are Riemann integrable on interval $[0,1]$ such that $$\int_0^1f(x)\mathrm{d}x=\int_0^1g(x)\mathrm{d}x=1.$$ Show that there exists some intervals $I\subset[0,1]$ such that $$\int_If(x)\mathrm{d}x=\int_Ig(x)\mathrm{d}x=\frac{1}{2}.$$

In my textbook it used some method that I never came across before, which involved the application of something like "winding number". I was wondering whether there are some "elementary method" to solve this, I know some real analysis (I finished reading Rudin's RCA before Chapter 6) and I wonder if some opinions in measure theory or Lebesgue integral is useful.

I'm looking for any suggestions for this problem. Thanks in advance.

asked Oct 31 '23 at 15:14

MathLearner

2

Does this answer your question? Measurable functions with the same integral over a set Or this, or this. (All found by searching for $\int_I f = \int_I g = 1/2$ on approach0. – Jakob Streipel Oct 31 '23 at 15:47

$\int_0^1f=\int_0^1g=1$ implies $\int_If=\int_Ig=\frac{1}{2}$ on some interval $I\subset[0,1]$

0 Answers0