# Riemann Integrability

• May 15th 2008, 01:03 PM
namelessguy
Riemann Integrability
I have no idea how to prove this. Can someone help?
Given a real number x, denote by [x] the largest integer less than or equal to x. Prove that the function defined
$\displaystyle f(x)$ =$\displaystyle \sum_{n = 0}^{\infty}\frac{nx-[nx]}{2^n}$ is Riemann integrable on any interval [a,b]. Then compute
$\displaystyle \int_0^1 f(x)dx$
• May 15th 2008, 03:11 PM
NonCommAlg
Quote:

Originally Posted by namelessguy
I have no idea how to prove this. Can someone help?
Given a real number x, denote by [x] the largest integer less than or equal to x. Prove that the function defined
$\displaystyle f(x)$ =$\displaystyle \sum_{n = 0}^{\infty}\frac{nx-[nx]}{2^n}$ is Riemann integrable on any interval [a,b]. Then compute
$\displaystyle \int_0^1 f(x)dx$

for n = 0 the term in your series is 0. so i'll assume that n > 0. let $\displaystyle f_n(x)=\frac{nx - [nx]}{2^n}, \ n \geq 1.$

each $\displaystyle f_n$ is integrable on any interval [a, b], because it's continuous almost everywhere on the

interval. let $\displaystyle \sigma_n(x)=\sum_{j=1}^n f_j(x).$ so $\displaystyle \sigma_n$ is integrable on [a, b] for all n.

since $\displaystyle 0 \leq f_n(x) \leq \frac{1}{2^n}, \ \forall x,$ by Weierstrass test, $\displaystyle \{\sigma_n \}$ is uniformly convergent. so $\displaystyle f(x)=\lim_{n\to\infty} \sigma_n(x)$

is integrable on [a, b] and $\displaystyle \int_a^b f(x) dx = \lim_{n\to\infty} \int_a^b \sigma_n(x) dx.$ if a = 0 and b = 1, then this gives us:

$\displaystyle \int_0^1 f(x) dx = \lim_{n\to\infty} \int_0^1 \sigma_n(x) dx = \lim_{n\to\infty} \int_0^1 \sum_{j=1}^n f_j(x) \ dx=\lim_{n\to\infty} \sum_{j=1}^n \int_0^1 f_j(x) dx$

$\displaystyle =\lim_{n\to\infty} \sum_{j=1}^n \int_0^1 \frac{jx -[jx]}{2^j}dx=\lim_{n\to\infty} \sum_{j=1}^n \frac{1}{2^{j+1}}=\sum_{j=1}^{\infty} \frac{1}{2^{j+1}}=\frac{1}{2}. \ \ \ \square$
• May 15th 2008, 03:22 PM
ThePerfectHacker
Quote:

Originally Posted by namelessguy
I have no idea how to prove this. Can someone help?
Given a real number x, denote by [x] the largest integer less than or equal to x. Prove that the function defined
$\displaystyle f(x)$ =$\displaystyle \sum_{n = 0}^{\infty}\frac{nx-[nx]}{2^n}$ is Riemann integrable on any interval [a,b]. Then compute
$\displaystyle \int_0^1 f(x)dx$

I will do the special case $\displaystyle [a,b]=[0,1]$. Define a sequence of functions $\displaystyle f_n:[0,1]\mapsto \mathbb{R}$ as $\displaystyle f_n(x) = \tfrac{1}{2^n}(nx-[nx])$. Notice that $\displaystyle |f_n(x)| \leq \tfrac{1}{2^n}$. Since $\displaystyle \sum _{n=0}^{\infty}\tfrac{1}{2^n} < \infty$ it follows by the Weierstrass test that the series of functions $\displaystyle \sum_{n=0}^{\infty} f_n(x)$ converges uniformly to a function $\displaystyle f(x)$. Each function $\displaystyle f_n(x)$ is discontinous at $\displaystyle \tfrac{k}{n}$ for $\displaystyle 1\leq k<n$. This implies that $\displaystyle f$ is integrable and $\displaystyle \int_0^1 f(x) dx = \sum_{n=0}^{\infty}\int_0^1 f_n(x) dx$. It remains to compute $\displaystyle \int_0^1 f_n(x) dx$. The graph of $\displaystyle f_n$ consists of a $\displaystyle n$ congruent right-angles triangles each having height $\displaystyle \tfrac{1}{2^n}$ and with base width $\displaystyle 1/n$. Thus, $\displaystyle \int_0^1 f_n(x) dx = \frac{1}{2} n\cdot \frac{1}{n} \cdot \frac{1}{2^n} = \frac{1}{2^{n+1}}$. I might have missed maybe $\displaystyle \pm 1$ on the exponent, but you get the idea. Now this is a geometric series which sums easily.

EDIT: Got beaten in the response by a fellow algebraist. (Hi)