Are Sigma-Finite Measures Always Lebesgue Measurable?

[island-boy]

hello, I need some help with 2 problesm involving sigma-finite measures.

*note* a set X is sigma-finite if X can be written as a compact union of subsets, i.e. X = Union of Xi for i = 1,2,3,...n, and the measure of each Xi is finite.

q1)Given a measure space (X, M, m), let f be lebesgue measurable such that f(x)>0 for every x element of X, show that m is sigma-finite.
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I think I may have solved this. Please see if I did anything wrong.

Let X = union of arbitrary disjoint subsets Xi where i =1,2,3...n

since f is lebesgue measurable, then f(x)m(X) is finite, hence m(X) is finite.
Since m(X) = m (UXi) = summation m(Xi) is finite. Hence m(Xi) is finite.
hence m is sigma finite. Is this correct?

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q2) suppose f is sigma-finite, constrauct a lebesgue measurable function s.t. f(x) > 0 for every x element of X abd the integral of f(x)dx over X = 1.

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I'm thinking this is how to solve this.
since the integral of f(x)dx over X = 1 then
f(x)m(X) = 1
and
f(x) = 1/m(X)...this f(x) is finite, hence it is lebesgue meaurable.

I'm thinking I need to get rid of the m(x) part, but I have no idea how. help please?

 

[island-boy]

anyone? help please. thanks!