- #1

Faris

- 1

- 0

First, I want to thank you people for the help.

**1. Define a function f by setting f(x)=1 if x=1/m for some positive integer m (e.g. f(1)=1, f(1/2)=1, f(1/3)=1,...) and f(x)=0 otherwise. Show that f is integrable on the interval [0,1] by proving that the limit of the lower Riemann sums for the regular partitions P_n equals the limit of the corresponding upper Riemann sums.**

**2. Lower Riemann sum = L(f,P_n)**

Upper Riemann sum = U(f,P_n)

Upper Riemann sum = U(f,P_n)

**3. I've tried the following argument, but I don't think it's enough for a proof:**

L(f,P_n) = 0 because by taking n to infinity, the minimum values for f(x) in the delta x's are zeros, which makes L(f,P_n)=0.

U(f,P_n), however, is the problem here.

L(f,P_n) = 0 because by taking n to infinity, the minimum values for f(x) in the delta x's are zeros, which makes L(f,P_n)=0.

U(f,P_n), however, is the problem here.

Thanks in advance.