Proof of Fubini's Theorem: Integrating h(x,y)

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SUMMARY

The proof of Fubini's Theorem states that for sigma-finite measure spaces (X,B,u) and (Y,C,v), if functions f and g belong to L1 spaces, then the product function h(x,y) = f(x)g(y) is also in L1(XxY,BxC,uxv). The theorem asserts that the integral of the product can be expressed as the product of the integrals: ∫ h d(u×v) = ∫ f du ∫ g dv. The key to proving that h is in L1 lies in recognizing that h can be decomposed into the product of its marginal functions hx and hy, which are both integrable.

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What's the proof of this fundamental theorem?
Let (X,B,u) and (Y,C,v) be sigma finite measure spaces, Let f in L1(X,B,u) and g in L1(Y,C,v). Let h(x,y)=f(x)g(y).
Then, h is in L1(XxY,BxC,uxv) and
[tex]\int hd(u\times v)=\int fdu \int gdv[/tex]

should be an easy application of fubini,but i really have no idea to how work it out
 
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Can you at least prove that h is L^1?
 
i know the trick must has something to do with
hx=g(y)
hy=f(x)
but then h=hx*hy, why is it in L1?
 

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