Convergence in Sigma-Finite Measure

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Theorem

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

Let $\sequence {f_n}_{n \mathop \in \N}: f_n: X \to \R$ be a sequence of measurable functions.

Also, let $f, g: X \to \R$ be measurable functions.

Suppose that $f_n$ converges in measure to both $f$ and $g$ (in $\mu$).


Then $f$ and $g$ are equal $\mu$-almost everywhere.


Proof

As $\mu$ is $\sigma$-finite, there is an exhausting sequence $\sequence {E_m}_m$ in $\Sigma$ such that:

$\forall m : \map \mu {E_m} < \infty$

By triangle inequality:

$\forall x \in X :\quad \cmod {\map f x - \map g x} \le \cmod {\map f x - \map {f_n} x} + \cmod {\map {f_n} x - \map g x}$

For all $m, k \in \N_{>0}$, since:

$\ds \forall n \in \N :\quad \set {\cmod {f - g} > \frac 1 k } \subseteq \set {\cmod {f - f_n} > \frac 1 {2 k} } \cup \set {\cmod {f_n - g} > \frac 1 {2 k} }$

we have:

\(\ds \map \mu {\set {\cmod {f - g} > \frac 1 k} \cap E_m }\) \(\le\) \(\ds \map \mu {\paren {\set {\cmod {f - f_n} > \frac 1 {2 k} } \cup \set {\cmod {f_n - g} > \frac 1 {2 k} } } \cap E_m }\)
\(\ds \) \(=\) \(\ds \map \mu { \paren {\set {\cmod {f - f_n} > \frac 1 {2 k} } \cap E_m } \cup \paren {\set {\cmod {f_n - g} > \frac 1 {2 k} } \cap E_m } }\) Intersection Distributes over Union
\(\ds \) \(\le\) \(\ds \map \mu {\set {\cmod {f - f_n} > \frac 1 {2 k} } \cap E_m } + \map \mu {\set {\cmod {f_n - g} > \frac 1 {2 k} } \cap E_m }\) Measure is Subadditive
\(\ds \) \(\to\) \(\ds 0\) as $n \to \infty$ by hypothesis

Thus:

$\ds \forall m, k \in \N_{>0} : \map \mu {\set {\cmod {f - g} > \frac 1 k} \cap E_m} = 0$

By Monotone Convergence Theorem:

$\ds \forall k \in \N_{>0} : \map \mu {\set {\cmod {f - g} > \frac 1 k} } = \lim_{m \mathop \to \infty} \map \mu {\set {\cmod {f - g} > \frac 1 k} \cap E_m} = 0$


Now, observe:

$\ds \set {\cmod {f - g} > 0} = \bigcup_{k \mathop = 1}^\infty \set {\cmod {f - g} > \frac 1 k}$

Therefore:

\(\ds \map \mu {\set {\cmod {f - g} > 0} }\) \(=\) \(\ds \map \mu {\bigcup_{k\mathop = 1}^\infty \set {\cmod {f - g} > \frac 1 k} }\)
\(\ds \) \(\le\) \(\ds \sum_{k\mathop = 1}^\infty \map \mu {\set {\cmod {f - g} > \frac 1 k} }\) Measure is Subadditive
\(\ds \) \(=\) \(\ds 0\)

$\blacksquare$


Sources

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