Let f:[0,1]→R be a continuous function such that f(0)≠f(1). Define fn(x)=f(xn) then I want to prove that limn→∞∫10fn(x)dx=f(0).

Now, you can easily prove that the convergence is not uniform, so we can’t switch the sign of limit and the sign of the integral. I tried to do a bunch of things to solve this problem, for example if S(f_n,\sigma_N) is the upper sum of f_n with respect to the equispaced partition \sigma_N then i tried to prove that \lim_{n\rightarrow\infty}\lim_{N\rightarrow\infty}S(f_n,\sigma_n)=f(0), this is true if we can switch the limits, but I don’t know why we can. Could you help me please?

**Answer**

Divide the interval in two parts:

A small interval near 1 where the integral is small because it is the length of the intervall times a bound of the function.

A large interval where you can bound the value of the function to be near f(0) for large n.

**Attribution***Source : Link , Question Author : Mec , Answer Author : Phira*