Characterization of Essentially Bounded Functions

Theorem

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

Let $f : X \to \R$ be a $\Sigma$-measurable function.

The following statements are equivalent:

$(1) \quad$ $f$ is essentially bounded.

$(2) \quad$ There exists a bounded function $g : X \to \R$ such that $f = g$ $\mu$-almost everywhere.

Proof

$(1)$ implies $(2)$

Suppose that there exists $c \in \R$ such that:

$\map \mu {\set {x \in X : \size {\map f x} > c} } = 0$

Let:

$A = \set {x \in X : \size {\map f x} \le c}$

Define a function $g : X \to \overline \R$ by:

$\map g x = \begin{cases}\map f x & x \in A \\ 0 & x \not \in A\end{cases}$

Then:

$\size {\map g x} \le c$ for all $x \in X$.

So:

$g$ is bounded.

Then if $x \in X$ is such that $\map f x \ne \map g x$, we must have $x \in X \setminus A$.

But by hypothesis:

$\map \mu {X \setminus A} = \map \mu {\set {x \in X : \size {\map f x} > c} } = 0$

So:

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

$\Box$

$(2)$ implies $(1)$

Suppose that:

there exists a bounded function $g : X \to \R$ such that $f = g$ $\mu$-almost everywhere.

That is, there exists a $\mu$-null set $N \subseteq X$ such that:

if $x \in X$ is such that $\map f x \ne \map g x$ then $x \in N$.

Let:

$\ds M = \sup_{x \mathop \in X} \size {\map g x} < \infty$

Since:

$g \le M$

if $x \in X$ is such that:

$\map f x > M$

then:

$\map f x \ne \map g x$

so:

$x \in N$

So:

$\set {x \in X : \size {\map f x} > M} \subseteq N$

From Absolute Value of Measurable Function is Measurable, we have that:

$\size f$ is $\Sigma$-measurable.

So, from Characterization of Measurable Functions we have:

$\set {x \in X : \size {\map f x} > M}$ is $\Sigma$-measurable.

Then from Null Sets Closed under Subset, since $\map \mu N = 0$, we have:

$\map \mu {\set {x \in X : \size {\map f x} > M} } = 0$

That is:

$f$ is essentially bounded.

$\blacksquare$