Proving Measurability and Integrability of a Function on a Product Space

[Funky1981]

Homework Statement

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)

Homework 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

 

[micromass]

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

[tex]h = f\chi_A + g\chi_B[/tex]

Show that ##h## is measurable.

Use this to show (1).

 

[Funky1981]

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

[tex]h = f\chi_A + g\chi_B[/tex]

Show that ##h## is measurable.

Use this to show (1).

So if I take A to be 0<y<x<1, B to be 0<x≤y<1 and construct [tex]h = f\chi_A + g\chi_B[/tex]
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?

 

[micromass]

So if I take A to be 0<y<x<1, B to be 0<x≤y<1 and construct [tex]h = f\chi_A + g\chi_B[/tex]
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?

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.