Pointwise Exponentiation preserves A.E. Equality
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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.
$\blacksquare$