# Integration of a piecewise function using measures

• Jul 18th 2011, 01:18 PM
Integration of a piecewise function using measures
Problem: Define a function f by $\displaystyle f(x)= \begin{cases} x, \quad& 0 \le x \le 1 \\0, \quad& otherwise \end{cases}.$

Given that $\displaystyle \mu$ is the Lebesgue measure on $\displaystyle \mathbb {R}$ i.e. $\displaystyle \mu ((a,b)) = b-a$

Show that $\displaystyle \int f d \mu = \frac {1}{2}$

My proof so far:

Using the hints in the text, I'm defining a series of function $\displaystyle f_n$ that converges to $\displaystyle f$, with $\displaystyle \int f_n d \mu \rightarrow \int f d \mu$

For $\displaystyle x \in (0,1]$, let $\displaystyle f_n(x) = \frac { [2^nx]}{2^n}$ where $\displaystyle [x]$ a rounded down to the nearest integer.

$\displaystyle f_n(x)= \begin{cases} \frac { [2^nx]}{2^n}, \quad& x \in (0,1] \\ 0, \quad& otherwise \end{cases}.$

Then I have $\displaystyle \int f_n (x) d \mu = \sum ^n _{k=1} ( \frac {1}{2^k} ) \mu \{ ( \frac {1}{2^k} , \frac {1}{2 ^{k-1}} )$
$\displaystyle = \sum ^n _{k=1} \frac {1}{4^k}$
$\displaystyle \frac {1}{3}$

But the answer is suppose to be $\displaystyle \frac {1}{2}$! I'm guessing my integration of the series of function is wrong, but why?
• Jul 18th 2011, 01:36 PM
Siron
Re: Integration of a piecewise function using measures
For entering latex you have to use: [ tex ]... [ / tex ] not [ math ] ... [/ math ]