Almost All Horizontal Sections of Integrable Function are Integrable

Theorem
Let $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$ be $\sigma$-finite measure spaces.

Let $\struct {X \times Y, \Sigma_X \otimes \Sigma_Y, \mu \times \nu}$ be the product measure space of $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$.

Let $f: X \times Y \to \overline \R_{\ge 0}$ be a $\mu \times \nu$-integrable function.

Then:


 * $f^y$ is $\mu$-integrable for $\nu$-almost all $y \in Y$

where $f^y$ is the $y$-horizontal section of $f$.

Proof
From Horizontal Section of Measurable Function is Measurable, we have:


 * $f^y$ is $\Sigma_X$-measurable for each $y \in Y$.

From Function Measurable iff Positive and Negative Parts Measurable, we have:


 * $\paren {f^y}^+$ is $\Sigma_X$-measurable for each $y \in Y$

and:


 * $\paren {f^y}^-$ is $\Sigma_X$-measurable for each $y \in Y$.

Define a function $g : Y \to \overline \R$ by:


 * $\ds \map g y = \int \paren {f^y}^+ \rd \mu$

for each $y \in Y$.

Define $h : Y\to \overline \R$ by:


 * $\ds \map h y = \int \paren {f^y}^- \rd \mu$

for each $y \in Y$.

From Integral of Horizontal Section of Measurable Function gives Measurable Function, we have:


 * $g$ and $h$ are $\Sigma_Y$-measuarble.

From Positive Part of Horizontal Section of Function is Horizontal Section of Positive Part, we have:


 * $\paren {f^y}^+ = \paren {f^+}^y$

From Negative Part of Horizontal Section of Function is Horizontal Section of Negative Part, we have:


 * $\paren {f^y}^- = \paren {f^-}^y$

We then have:

Similarly:

Since $f$ is $\mu \times \nu$-integrable, we have:


 * $\ds \int_{X \times Y} f^+ \map \rd {\mu \times \nu} < \infty$

and:


 * $\ds \int_{X \times Y} f^- \map \rd {\mu \times \nu} < \infty$

So:


 * $g$ and $h$ are $\mu$-integrable.

From Integrable Function is A.E. Real-Valued, we therefore have:


 * $\ds \int \paren {f^y}^+ \rd \mu < \infty$

and:


 * $\ds \int \paren {f^y}^- \rd \mu < \infty$

for $\nu$-almost all $y \in Y$.

That is, there exists a $\nu$-null set $N_1 \subseteq Y$ such that whenever:


 * $\ds \int \paren {f^y}^+ \rd \mu = \infty$

we have $y \in N_1$.

Similarly, there exists a $\nu$-null set $N_2 \subseteq Y$ such that whenever:


 * $\ds \int \paren {f^y}^- \rd \mu = \infty$

we have $y \in N_2$.

If $f^y$ is $\mu$-integrable, we have either:


 * $\ds \int \paren {f^y}^+ \rd \mu < \infty$

or:


 * $\ds \int \paren {f^y}^- \rd \mu < \infty$

that is:


 * $y \in N_1$ or $y \in N_2$.

That is, if $f^y$ is not $\mu$-integrable, we have:


 * $y \in N_1 \cup N_2$

From Null Sets Closed under Countable Union, we have:


 * $N_1 \cup N_2$ is $\nu$-null.

So:


 * $f^y$ is $\mu$-integrable for $\nu$-almost all $y \in Y$.