Pointwise Exponentiation preserves A.E. Equality

Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space and let $p \in \hointr 0 \infty$.

Let $f, g : X \to \R$ be real-valued $\Sigma$-measurable functions such that:

$\size f = \size g$ $\mu$-almost everywhere.

Then:

$\size f^p = \size g^p$ $\mu$-almost everywhere.

Proof

Since:

$\size f = \size g$ $\mu$-almost everywhere

there exists a $\mu$-null set $N$ such that:

if $\size {\map f x} \ne \size {\map g x}$ then $x \in N$.

If $\size {\map f x} = \size {\map g x}$, then $\size {\map f x}^p = \size {\map g x}^p$.

So by the Rule of Transposition, we have if $\size {\map f x}^p \ne \size {\map g x}^p$ then $\size {\map f x} \ne \size {\map g x}$.

So, if $x \in X$ has $\size {\map f x}^p \ne \size {\map g x}^p$ then $x \in N$.

So:

$\size f^p = \size g^p$ $\mu$-almost everywhere.

$\blacksquare$