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