Integral on L-1 Space is Well-Defined

Theorem

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

Let $\map {L^1} {X, \Sigma, \mu}$ be the $L^1$ space of $\struct {X, \Sigma, \mu}$.

Let $E \in \map {L^1} {X, \Sigma, \mu}$.

Then the $\mu$-integral of $E$ is well-defined.

Proof

Let $E = \eqclass f \sim$.

We first establish that:

$\ds \int f \rd \mu$

is a well-understood real number.

We then will show that the $\mu$-integral of $E$ is independent of the choice of representative for $E$.

Since $E \in \map {L^1} {X, \Sigma, \mu}$, we have that $f \in \map {\LL^1} {X, \Sigma, \mu}$.

Then:

$\ds \int \size f \rd \mu < \infty$

So, from Characterization of Integrable Functions, we have:

$\ds \int f \rd \mu$ is well-defined and exists as a finite number.

Suppose now that $\eqclass f \sim = \eqclass g \sim$.

From Equivalence Class Equivalent Statements:

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

from the definition of the $\mu$-almost everywhere relation.

From A.E. Equal Positive Measurable Functions have Equal Integrals: Corollary 1, we have:

$\ds \int f \rd \mu = \int g \rd \mu$

So:

$\ds \int E \rd \mu$

is independent of the choice of representative for $E$.

$\blacksquare$