Fubini's theorem for integrable functions.

Question

I have gone through the proof of Fubini's theorem for non-negative measurable functions from the book An Introduction to Measure and Integration by Inder K Rana. The satement of the theorem is as follows $:$

Theorem $1$ $:$ Let $(X \times Y, \mathcal A \otimes \mathcal B, \mu \times \nu)$ be the product measure space induced by the $\sigma$-finite measure spaces $(X,\mathcal A, \mu)$ and $(Y,\mathcal B, \nu).$ Then for any non-negative $\mathcal A \otimes \mathcal B$- measurable function $f,$ the following staements hold $:$

$($i$)$ For any $x_0 \in X,y_0 \in Y$ the maps $x \longmapsto f(x,y_0)$ and $y \longmapsto f(x_0,y)$ are $\mathcal A$-measurable and $\mathcal B$-measurable respectively.

$($ii$)$ The map $x \longmapsto \displaystyle {\int_{Y}} f(x,y)\ d\nu(y)$ is $\mathcal A$-measurable and the map $y \longmapsto \displaystyle {\int_{X}} f(x,y)\ d\mu(x)$ is $\mathcal B$-measurable.

$($iii$)$ $\displaystyle {\int_{X}} \left ( \displaystyle {\int_{Y}} f(x,y)\ d\nu(y) \right ) d\mu(x) = \displaystyle {\int_{Y}} \left ( \displaystyle {\int_{X}} f(x,y)\ d\mu(x) \right ) d\nu(y) = \displaystyle {\int_{X \times Y}} f(x,y)\ d(\mu \times \nu) (x,y).$

The general version of the above theorem states as follows $:$

Theorem $2$ $:$ Let $(X \times Y, \mathcal A \otimes \mathcal B, \mu \times \nu)$ be the product measure space induced by the $\sigma$-finite measure spaces $(X,\mathcal A, \mu)$ and $(Y,\mathcal B, \nu).$ Then for any $f \in L_1 (\mu \times \nu),$ the following staements hold $:$

$($i$)$ The maps $x \longmapsto f(x,y)$ and $y \longmapsto f(x,y)$ are $\mu$-integrable a.e. $y(\nu)$ and $\nu$-integrable a.e. $x(\mu)$ respectively.

$($ii$)$ The map $x \longmapsto \displaystyle {\int_{Y}} f(x,y)\ d\nu(y)$ is $\mu$-integrable a.e. $x(\mu)$ and the map $y \longmapsto \displaystyle {\int_{X}} f(x,y)\ d\mu(x)$ is $\nu$-integrable a.e. $y(\nu).$

$($iii$)$ $\displaystyle {\int_{X}} \left ( \displaystyle {\int_{Y}} f(x,y)\ d\nu(y) \right ) d\mu(x) = \displaystyle {\int_{Y}} \left ( \displaystyle {\int_{X}} f(x,y)\ d\mu(x) \right ) d\nu(y) = \displaystyle {\int_{X \times Y}} f(x,y)\ d(\mu \times \nu) (x,y).$

I tried to prove the above theorem with the help of Theorem $1.$ Here's what I did $:$

My attempt $:$ Let $f^+$ and $f^-$ be the positive and the negative part of the function $f$ respectively. Since $f \in L_1(\mu \times \nu),$ $f^+$ and $f^-$ are both non-negative $\mathcal A \otimes \mathcal B$-measurable functions. Applying Theorem $1$ $($iii$)$ to $f^+$ and $f^{-}$ we have

\begin{align*}\displaystyle {\int_{X}} \left ( \displaystyle {\int_{Y}} f^+(x,y)\ d\nu(y) \right ) d\mu(x) = \displaystyle {\int_{Y}} \left ( \displaystyle {\int_{X}} f^+(x,y)\ d\mu(x) \right ) d\nu(y) & = \displaystyle {\int_{X \times Y}} f^+(x,y)\ d(\mu \times \nu) (x,y) \\ & \leq \displaystyle {\int_{X \times Y}} |f(x,y)|\ d(\mu \times \nu) < +\infty. \end{align*}

\begin{align*}\displaystyle {\int_{X}} \left ( \displaystyle {\int_{Y}} f^-(x,y)\ d\nu(y) \right ) d\mu(x) = \displaystyle {\int_{Y}} \left ( \displaystyle {\int_{X}} f^-(x,y)\ d\mu(x) \right ) d\nu(y) & = \displaystyle {\int_{X \times Y}} f^-(x,y)\ d(\mu \times \nu) (x,y) \\ & \leq \displaystyle {\int_{X \times Y}} |f(x,y)|\ d(\mu \times \nu) < +\infty. \end{align*}

This shows that the map $x \longmapsto \displaystyle {\int_Y} f^+(x,y)\ d\nu(y)$ is $\mu$-integrable, the map $y \longmapsto \displaystyle {\int_X} f^+(x,y)\ d\mu(x)$ is $\nu$-integrable, the map $x \longmapsto \displaystyle {\int_Y} f^-(x,y)\ d\nu(y)$ is $\mu$-integrable and the map $y \longmapsto \displaystyle {\int_X} f^-(x,y)\ d\mu(x)$ is $\nu$-integrable.

So the map $y \longmapsto f^+(x,y)$ is $\nu$-integrable a.e. $x(\mu)$ and the map $y \longmapsto f^-(x,y)$ is $\nu$-integrable a.e. $x(\mu).$ Hence $y \longmapsto f(x,y)$ is $\nu$-integrable a.e. $x(\mu).$ Similarly, the map $x \longmapsto f^+(x,y)$ is $\mu$-integrable a.e. $y(\nu)$ and the map $x \longmapsto f^-(x,y)$ is $\mu$-integrable a.e. $y(\nu).$ Hence $x \longmapsto f(x,y)$ is $\mu$-integrable a.e. $y(\nu).$ This proves $($i$).$

Since $f \in L_1(\mu \times \nu)$ it follows that \begin{align*} \int_{X \times Y} f(x,y)\ d(\mu \times \nu) (x,y) & = \int_{X \times Y} f^+(x,y)\ d(\mu \times \nu) (x,y) - \int_{X \times Y} f^-(x,y)\ d(\mu \times \nu) (x,y) \\ & = \int_X \left ( \int_{Y} f^+(x,y)\ d{\nu(y)} \right ) d{\mu}(x) - \int_X \left ( \int_{Y} f^-(x,y)\ d{\nu(y)} \right ) d{\mu}(x) \end{align*}

Now how do I proceed? Any help will be highly appreciated.

Thanks in advance.

Answer 1

The assertion of Fubini's theorem for any integrable function what has been made in the book An Introduction to Measure and Integration by Inder K Rana is not correct. It should be the following $:$

Theorem (Fubini) $:$ Let $(X, \mathcal A, \mu)$ and $(Y,\mathcal B, \nu)$ be two complete $\sigma$-finite measure spaces. Let $(X \times Y,\mathcal A \otimes \mathcal B,\mu \times \nu)$ be the product measure space induced by $(X,\mathcal A, \mu)$ and $(Y,\mathcal B, \nu).$ Let $f \in L_1(\mu \times \nu).$ Then there exist $g \in L_1(\mu)$ and $h \in L_1(\nu)$ such that $$\int_{X \times Y} f\ d(\mu \times \nu) = \int_X g\ d\mu = \int_Y h\ d\nu.$$

Let us begin the proof from the last equality what I obtained i.e. \begin{align*} \int_{X \times Y} f(x,y)\ d(\mu \times \nu) (x,y) & = \int_X \left ( \int_{Y} f^+(x,y)\ d{\nu(y)} \right ) d{\mu}(x) - \int_X \left ( \int_{Y} f^-(x,y)\ d{\nu(y)} \right ) d{\mu}(x)\ \ \ \ {\label \equation (1)}\end{align*}

Let \begin{align*} E : & = \left \{x \in X\ \bigg |\ \int_Y f^+(x,y)\ d\nu(y) < +\infty \right \} \\ F : & = \left \{x \in X\ \bigg |\ \int_Y f^-(x,y)\ d\nu(y) < +\infty \right \} \end{align*} Since the maps $x \longmapsto \displaystyle {\int_Y} f^+(x,y)\ d\nu(y)$ and $x \longmapsto \displaystyle {\int_Y} f^-(x,y)\ d\nu(y)$ are both $\mu$-integrable it follows that $\mu (E^c) = \mu(F^c) = 0.$ Define a function $g^+ : X \longrightarrow \Bbb R$ defined by $$g^+(x) = \left ( \displaystyle {\int_Y} f^+(x,y)\ d\nu(y) \right ) \chi_E (x),\ x \in X$$ and a function $g^- : X \longrightarrow \Bbb R$ defined by $$g^-(x) = \left ( \displaystyle {\int_Y} f^-(x,y)\ d\nu(y) \right ) \chi_F (x),\ x \in X$$ Then clearly $g^+(x),g^-(x) < +\infty,\ $ for all $x \in X.$ Moreover \begin{align*} g^+(x) & = \displaystyle {\int_Y} f^+(x,y)\ d\nu(y) ,\ \text{for a.e.}\ x(\mu) \\ g^-(x) & = \displaystyle {\int_Y} f^-(x,y)\ d\nu(y) ,\ \text{for a.e.}\ x(\mu) \end{align*} Let $g : = g^+ - g^-.$ Since the maps $x \longmapsto \displaystyle {\int_Y} f^+(x,y)\ d\nu(y)$ and $x \longmapsto \displaystyle {\int_Y} f^-(x,y)\ d\nu(y)$ are both $\mu$-integrable and $(X,\mathcal A,\mu)$ is a complete measure space it follows that $g^+,g^-,g \in L_1(\mu)$ and we have the following equality \begin{align*} \int_X g^+\ d\mu & = \int_X \left (\int_Y f^+(x,y)\ d\nu(y) \right ) d\mu(x) \\ \int_X g^-\ d\mu & = \int_X \left (\int_Y f^-(x,y)\ d\nu(y) \right ) d\mu(x) \\ \int_X g\ d\mu & = \int_X g^+\ d\mu - \int_X g^-\ d\mu \end{align*} From the above three equalities it follows that $$\int_X \left (\int_Y f^+(x,y)\ d\nu(y) \right ) d\mu(x) - \int_X \left (\int_Y f^-(x,y)\ d\nu(y) \right ) d\mu(x) = \int_X g\ d\mu.$$

Now from $(1)$ it follows that $$\int_{X \times Y} f\ d(\mu \times \nu) = \int_X g\ d\mu.$$

Similarly by observing that \begin{align*} \int_{X \times Y} f(x,y)\ d(\mu \times \nu) (x,y) & = \int_Y \left ( \int_{X} f^+(x,y)\ d{\mu(x)} \right ) d{\nu}(y) - \int_Y \left ( \int_{X} f^-(x,y)\ d{\mu(x)} \right ) d{\nu}(y) \end{align*} and by exploiting the completeness of the measure space $(Y,\mathcal B,\nu)$ we can find out $h \in L_1(\nu)$ such that $$\int_{X \times Y} f\ d(\mu \times \nu) = \int_Y h\ d\nu.$$

This completes the proof.

QED

Answer 2

I have the following canonical detailed answer on Fubini’s theorem. My source is “Elements of the theory of functions and functional analysis” by A.N. Kolmogorov and S.V. Fomin (Vol. 2, Graylock Press, Albany, N.Y. 1961). It is translated from the first (1960) Russian edition by Hyman Kamel and Horace Komm. Also I checked the formulations with the forth (1975) revised Russian edition, where it looks essentially the same.