# Measurable and integrable

1. Apr 21, 2014

### Funky1981

1. The problem statement, all variables and given/known data
Let f : (0,1) —>R be measurable( w.r.t. Lebesgue measure) function in L1((0,1)). Define the function g on (0,1)× (0,1) by

g(x,y)=f(x)/x if 0<y<x<1
g(x,y)=0 if 0<x≤y<1

Prove:
1) g is measurable function (w.r.t. Lebesgue measure in the prodcut (0,1)× (0,1)

2)g is integrable in (0,1)× (0,1)

2. Relevant equations

I tried to think about the product space but it is not really obvious for me how to begin, Can anyone give me suggestions

2. Apr 21, 2014

### micromass

Staff Emeritus
Start by showing this:

Let $(\Omega,\mathcal{B})$ is a set with $\sigma$-algebra.
Let $A,B\in \mathcal{B}$ disjoint, define $\chi_A$ and $\chi_B$ the indicator functions: http://en.wikipedia.org/wiki/Indicator_function#Definition
Let $f:\Omega\rightarrow \mathbb{R}$ and $g:\Omega\rightarrow \mathbb{R}$ be measurable.

Define $h:\Omega\rightarrow \mathbb{R}$ by

$$h = f\chi_A + g\chi_B$$

Show that $h$ is measurable.

Use this to show (1).

3. Apr 21, 2014

### Funky1981

So if I take A to be 0<y<x<1, B to be 0<x≤y<1 and construct $$h = f\chi_A + g\chi_B$$
consider set {x| h(x)> a} if it is in A then we have h= f then f measurable implies h measurable, is my construction right?

4. Apr 21, 2014

### micromass

Staff Emeritus
In my post, we had $A\in \Omega$ and $f$ had $\Omega$ as domain.
Now, here is the slight technical issue that $A\subseteq (0,1)^2$, but this is not the domain of $f$. So you need to solve this.