Pointwise Addition preserves A.E. Equality
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Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g, F, G : X \to \overline \R$ be functions with:
- $f = F$ $\mu$-almost everywhere
and:
- $g = G$ $\mu$-almost everywhere
and the pointwise sums $f + g$ and $F + G$ well-defined.
Then:
- $f + g = F + G$ $\mu$-almost everywhere.
Proof
Since:
- $f = F$ $\mu$-almost everywhere
there exists a $\mu$-null set $N_1 \subseteq X$ such that:
- if $x \in X$ has $\map f x \ne \map F x$ then $x \in N_1$.
Since:
- $g = G$ $\mu$-almost everywhere
there exists a $\mu$-null set $N_2 \subseteq X$ such that:
- if $x \in X$ has $\map G x \ne \map G x$ then $x \in N_2$
Note that if $x \in X$ is such that:
- $\map f x = \map F x$
and:
- $\map g x = \map G x$
then:
- $\map f x + \map g x = \map F x + \map G x$
whenever this sum is well-defined.
By the Rule of Transposition, we therefore have:
- if $x \in X$ has $\map f x + \map g x \ne \map F x + \map G x$ then $\map f x \ne \map F x$ or $\map g x \ne \map G x$.
That is, whenever $x \in X$ has $\map f x + \map g x \ne \map F x + \map G x$, 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:
- $f + g = F + G$ $\mu$-almost everywhere.
$\blacksquare$