L-2 Inner Product is Well-Defined

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Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $\map {\LL^2} {X, \Sigma, \mu}$ be the Lebesgue $2$-space of $\struct {X, \Sigma, \mu}$.

Let $\sim$ be the $\mu$-almost-everywhere equality relation on $\map {\LL^2} {X, \Sigma, \mu}$.

Let $\map {L^2} {X, \Sigma, \mu}$ be the $L^p$ space on $\struct {X, \Sigma, \mu}$.

Let $E, F \in \map {L^2} {X, \Sigma, \mu}$.


Then the $L^2$ inner product $\innerprod E F$ is well-defined.


Proof

Let $E = \eqclass f \sim$ and $F = \eqclass g \sim$ where $\sim$ is the $\mu$-almost everywhere equality relation.

We aim to show that:

$\ds \int \paren {f \cdot g} \rd \mu$

exists as a real number and is independent of the choice of representatives $f$ and $g$ of $E$ and $F$.

From the definition of the $L^2$ space, we have $f, g \in \map {\LL^2} {X, \Sigma, \mu}$.

From Hölder's Inequality for Integrals, we have:

$f \cdot g \in \map {\LL^1} {X, \Sigma, \mu}$

So the $\mu$-integral of $f \cdot g$ is well-defined and exists as a real number.

Now suppose that:

$\eqclass {f_1} \sim = \eqclass {f_2} \sim = E$

and:

$\eqclass {g_1} \sim = \eqclass {g_2} \sim = F$

From Equivalence Class Equivalent Statements, we have:

$f_1 \sim f_2$

and:

$g_1 \sim g_2$

That is:

$f_1 = f_2$ $\mu$-almost everywhere

and:

$g_1 = g_2$ $\mu$-almost everywhere.

From Pointwise Multiplication preserves A.E. Equality, we have:

$f_1 \cdot g_1 = f_2 \cdot g_2$ $\mu$-almost everywhere.

From A.E. Equal Positive Measurable Functions have Equal Integrals, we have:

$\ds \int \paren {f_1 \cdot g_1} \rd \mu = \int \paren {f_2 \cdot g_2} \rd \mu$

as desired.

$\blacksquare$