User:Caliburn/s/mt/Linear Combination of A.E. Equal Functions

Theorem

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

Let $f, g, h, k : X \to \overline \R$ be functions such that:

$f = h$ $\mu$-almost everywhere

and:

$g = k$ $\mu$-almost everywhere.

Let $\alpha, \beta \in \overline \R$.

Then:

$\alpha f + \beta g = \alpha h + \beta k$ $\mu$-almost everywhere.

Proof

Since $f = h$ $\mu$-almost everywhere, there exists a $\mu$-null set $N_1 \subseteq X$ such that:

if $x \in X$ has $\map f x \ne \map h x$, then $x \in N_1$.

Since $g = k$ $\mu$-almost everywhere, there exists a $\mu$-null set $N_2 \subseteq X$ such that:

if $x \in X$ has $\map g x \ne \map k x$, then $x \in N_2$.

Note that if $x \in X$ is such that $\map f x = \map h x$ and $\map g x = \map k x$, then:

$\alpha \map f x + \beta \map g x = \alpha \map h x + \beta \map k x$

So if $x \in X$ has:

$\alpha \map f x + \beta \map g x \ne \alpha \map h x + \beta \map k x$

then:

either $\map f x \ne \map h x$ or $\map g x \ne \map k x$.

That is:

either $x \in N_1$ or $x \in N_2$.

From the definition of set union, this is equivalent to:

$x \in N_1 \cup N_2$

From Null Sets Closed under Countable Union, we have:

$N_1 \cup N_2$ is $\mu$-null.

That is:

if $x \in X$ has $\alpha \map f x + \beta \map g x \ne \alpha \map h x + \beta \map k x$ then $x \in N_1 \cup N_2$, which is a $\mu$-null set.

So we have:

$\alpha f + \beta g = \alpha h + \beta k$ $\mu$-almost everywhere.