proof of Fubini’s theorem for the Lebesgue integral

Let ${\mu }_x$ and ${\mu }_y$ be measures on $X$ and $Y$ respectively, let $\mu$ be the product measure ${\mu }_x\otimes {\mu }_y$, and let $f(x,y)$ be $\mu$-integrable on $A\subset X\times Y$. Then

$${\int }_{A}f(x,y)𝑑\mu ={\int }_X\left({\int }_{A_x}f(x,y)𝑑{\mu }_y\right)𝑑{\mu }_x={\int }_Y\left({\int }_{A_y}f(x,y)𝑑{\mu }_x\right)𝑑{\mu }_y$$

where

$$A_x=\{y\mid (x,y)\in A\},A_y=\{x\mid (x,y)\in A\}$$

Proof: Assume for now that $f(x,y)\ge 0$. Consider the set

$$U=X\times Y\times ℝ$$

equipped with the measure

$${\mu }_u={\mu }_x\otimes {\mu }_y\otimes {\mu }^1=\mu \otimes {\mu }^1={\mu }_x\otimes \lambda$$

where ${\mu }^1$ is ordinary Lebesgue measure and $\lambda ={\mu }_y\otimes {\mu }^1$. Also consider the set $W\subset U$ defined by

$$W=\{(x,y,z)\mid (x,y)\in A,0\le z\le f(x,y)\}$$

Then

$${\mu }_u\left(W\right)={\int }_{A}f(x,y)𝑑\mu$$

And

$${\mu }_u\left(W\right)={\int }_X\lambda \left(W_x\right)𝑑{\mu }_x$$

where

$$W_x=\{(y,z)\mid (x,y,z)\in W\}$$

However, we also have that

$$\lambda \left(W_x\right)={\int }_{A_x}f(x,y)𝑑{\mu }_y$$

Combining the last three equations gives us Fubini’s theorem. To remove the restriction that $f(x,y)$ be nonnegative, write $f$ as

$$f(x,y)=f^{+}(x,y)-f^{-}(x,y)$$

where

$$f^{+}(x,y)=\frac{|f(x,y)|+f(x,y)}{2},f^{-}(x,y)=\frac{|f(x,y)|-f(x,y)}{2}$$

are both nonnegative.

Title	proof of Fubini’s theorem for the Lebesgue integral
Canonical name	ProofOfFubinisTheoremForTheLebesgueIntegral
Date of creation	2013-03-22 15:21:52
Last modified on	2013-03-22 15:21:52
Owner	azdbacks4234 (14155)
Last modified by	azdbacks4234 (14155)
Numerical id	4
Author	azdbacks4234 (14155)
Entry type	Proof
Classification	msc 28A35