Functions A.E. Equal iff Positive and Negative Parts A.E. Equal

Theorem

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

Let $f, g : X \to \overline \R$ be extended real-valued functions.

Then $f = g$ $\mu$-almost everywhere if and only if:

$f^+ = g^+$ $\mu$-almost everywhere and $f^- = g^-$ $\mu$-almost everywhere.

where:

$f^+$ and $g^+$ denote the positive parts of $f$ and $g$ respectively

$f^-$ and $g^-$ denote the negative parts of $f$ and $g$ respectively.

Proof

Necessary Condition

Suppose that $f = g$ $\mu$-almost everywhere.

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

if $x \in X$ has $\map f x \ne \map g x$, then $x \in N$.

From the definition of the positive part, we have:

$\map {f^+} x = \max \set {\map f x, 0}$

and:

$\map {g^+} x = \max \set {\map g x, 0}$

Clearly if $\map f x = \map g x$, we then have $\map {f^+} x = \map {g^+} x$.

So from Proof by Contraposition, if $\map {f^+} x \ne \map {g^+} x$ for $x \in X$, we have that $\map f x \ne \map g x$.

That is, if $\map {f^+} x \ne \map {g^+} x$, we have $x \in N$, which is $\mu$-null set.

So:

$f^+ = g^+$ $\mu$-almost everywhere.

From the definition of the negative part, we have:

$\map {f^-} x = -\min \set {\map f x, 0}$

and:

$\map {g^-} x = -\min \set {\map g x, 0}$

As before, by contraposition, if $\map {f^-} x \ne \map {g^-} x$ for $x \in X$, we have that $\map f x \ne \map g x$.

So, if $\map {f^-} x \ne \map {g^-} x$, then $x \in N$.

Again, $N$ is $\mu$-null so we also obtain:

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

$\Box$

Sufficient Condition

Suppose that:

$f^+ = g^+$ $\mu$-almost everywhere.

and:

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

Then there exists a $\mu$-null set $N_1 \subseteq X$ such that:

if $x \in X$ is such that $\map {f^+} x \ne \map {g^+} x$, we have $x \in N_1$.

There also exists a $\mu$-null set $N_2 \subseteq X$ such that:

if $x \in X$ is such that $\map {f^-} x \ne \map {g^-} x$, we have $x \in N_2$.

From Difference of Positive and Negative Parts, we have:

$f = f^+ - f^-$

and:

$g = g^+ - g^-$

If $x \in X$ is such that $\map {f^+} x = \map {g^+} x$ and $\map {f^-} x = \map {g^-} x$, we therefore have $\map f x = \map g x$.

So by contraposition, if $\map f x \ne \map g x$, then $\map {f^+} x \ne \map {g^+} x$ or $\map {f^-} x \ne \map {g^-} x$.

Now let $x \in X$ be such that $\map f x \ne \map g x$.

Then either:

$\map {f^+} x \ne \map {g^+} x$

in which case, $x \in N_1$, or:

$\map {f^-} x \ne \map {g^-} x$

in which case $x \in N_2$.

So if $x \in X$ is such that $\map f x \ne \map g x$, we have:

$x \in N_1 \cup N_2$

From Null Sets Closed under Countable Union, we have that:

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

So:

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

$\blacksquare$