Pointwise Sum of Measurable Functions is Measurable

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Theorem

Let $\struct {X, \Sigma}$ 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.


General Result

Let $\sequence {\alpha_n}_{n \mathop \in \N}$ be a sequence of real numbers.

Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence of $\Sigma$-measurable functions $f_n : X \to \overline \R$ such that:

for each $N \in \N$ and $x \in X$, the sum $\ds \sum_{n \mathop = 1}^N \alpha_n \map {f_n} x$ is well-defined.


Then:

$\ds \sum_{n \mathop = 1}^N \alpha_n f_n$ is $\Sigma$-measurable.


Proof

By Measurable Function is Pointwise Limit of Simple Functions, we find sequences $\sequence {f_n}_{n \mathop \in \N}, \sequence {g_n}_{n \mathop \in \N}$ such that:

$\ds f = \lim_{n \mathop \to \infty} f_n$
$\ds g = \lim_{n \mathop \to \infty} g_n$

where the limits are pointwise.

It follows that for all $x \in X$:

$\map f x + \map g x = \ds \lim_{n \mathop \to \infty} \map {f_n} x + \map {g_n} x$



so that we have the pointwise limit:

$\ds f + g = \lim_{n \mathop \to \infty} f_n + g_n$

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.

$\blacksquare$


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