Pointwise Sum of Measurable Functions is Measurable
Theorem
Let $\left({X, \Sigma}\right)$ be a measurable space.
Let $f, g: X \to \overline{\R}$ be $\Sigma$-measurable functions.
Assume that the pointwise sum $f + g: X \to \overline{\R}$ is well-defined.
Then $f + g$ is a $\Sigma$-measurable function.
Proof
By Measurable Function Pointwise Limit of Simple Functions, we find sequences $\left({f_n}\right)_{n \in \N}, \left({g_n}\right)_{n \in \N}$ such that:
- $\displaystyle f = \lim_{n \to \infty} f_n$
- $\displaystyle g = \lim_{n \to \infty} g_n$
where the limits are pointwise.
It follows that for all $x \in X$:
- $f \left({x}\right) + g \left({x}\right) = \displaystyle \lim_{n \to \infty} f_n \left({x}\right) + g_n \left({x}\right)$
There needs to be a bunch of results establishing such equalities of convergence in function spaces
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so that we have the pointwise limit:
- $\displaystyle f + g = \lim_{n \to \infty} f_n + g_n$
By Pointwise Sum of Simple Functions is Simple Function, $f + g$ is a pointwise limit of simple functions.
By Simple Function is Measurable, $f + g$ is a pointwise limit of measurable functions.
Hence $f + g$ is measurable, by Pointwise Limit of Measurable Functions is Measurable.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $8.10$