# 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 $f(x,y)\geq 0$. Consider the set

 $U=X\times Y\times\mathbb{R}$

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

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

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 ProofOfFubinisTheoremForTheLebesgueIntegral 2013-03-22 15:21:52 2013-03-22 15:21:52 azdbacks4234 (14155) azdbacks4234 (14155) 4 azdbacks4234 (14155) Proof msc 28A35