Pointwise Addition preserves A.E. Equality

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.