Almost All Vertical Sections of Integrable Function are Integrable

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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:

$f_x$ is $\nu$-integrable for $\mu$-almost all $x \in X$

where $x$ is the $x$-vertical section of function.


Proof

From Vertical Section of Measurable Function is Measurable, we have:

$f_x$ is $\Sigma_Y$-measurable for each $x \in X$.

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

$\paren {f_x}^+$ is $\Sigma_Y$-measurable for each $x \in X$

and:

$\paren {f_x}^-$ is $\Sigma_Y$-measurable for each $x \in X$.

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

$\ds \map g x = \int \paren {f_x}^+ \rd \nu$

for each $x \in X$.

Define $h : X \to \overline \R$ by:

$\ds \map h x = \int \paren {f_x}^- \rd \nu$

for each $x \in X$.

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

$g$ and $h$ are $\Sigma_X$-measuarble.

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

$\paren {f_x}^+ = \paren {f^+}_x$

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

$\paren {f_x}^- = \paren {f^-}_x$

We then have:

\(\ds \int g \rd \mu\) \(=\) \(\ds \int_X \paren {\int_Y \paren {f_x}^+ \rd \nu} \rd \mu\)
\(\ds \) \(=\) \(\ds \int_X \paren {\int_Y \paren {f^+}_x \rd \nu} \rd \mu\)
\(\ds \) \(=\) \(\ds \int_{X \times Y} f^+ \map \rd {\mu \times \nu}\) Tonelli's Theorem

Similarly:

\(\ds \int h \rd \mu\) \(=\) \(\ds \int_X \paren {\int_Y \paren {f_x}^- \rd \nu} \rd \mu\)
\(\ds \) \(=\) \(\ds \int_X \paren {\int_Y \paren {f^-}_x \rd \nu} \rd \mu\)
\(\ds \) \(=\) \(\ds \int_{X \times Y} f^- \map \rd {\mu \times \nu}\) Tonelli's Theorem

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:

$\ds \int \paren {f_x}^+ \rd \nu < \infty$

and:

$\ds \int \paren {f_x}^- \rd \nu < \infty$

for $\mu$-almost all $x \in X$.

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

$\ds \int \paren {f_x}^+ \rd \nu = \infty$

we have $x \in N_1$.

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

$\ds \int \paren {f_x}^- \rd \nu = \infty$

we have $x \in N_2$.

If $f_x$ is not $\nu$-integrable, we have either:

$\ds \int \paren {f_x}^+ \rd \nu = \infty$

or:

$\ds \int \paren {f_x}^- \rd \nu = \infty$

that is:

$x \in N_1$ or $x \in N_2$

That is, if $f_x$ is not $\nu$-integrable, we have:

$x \in N_1 \cup N_2$

From Null Sets Closed under Countable Union, we have:

$N_1 \cup N_2$ is $\mu$-null.

So:

$f_x$ is $\nu$-integrable for $\mu$-almost all $x \in X$.

$\blacksquare$


Sources

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