Pointwise Sum of Measurable Functions is Measurable/General Result

Theorem

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

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

We proceed by induction.

For all $N \in \N$ let $\map P N$ be the proposition:

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

Basis for Induction

From the construction of $\sequence {f_n}_{n \mathop \in \N}$ we have:

$f_1$ is $\Sigma$-measurable.

From Pointwise Scalar Multiple of Measurable Function is Measurable, we have:

$\alpha_1 f_1$ is is $\Sigma$-measurable.

This is precisely $\map P 1$.

This is our basis for the induction.

Induction Hypothesis

Now we need to show that, if $\map P N$ is true, where $N \ge 1$, then it logically follows that $\map P {N + 1}$ is true.

Our induction hypothesis is:

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

We aim to show that:

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

Induction Step

This is our induction step.

We have:

$\ds \sum_{n \mathop = 1}^{N + 1} \alpha_n f_n = \alpha_{N + 1} f_{N + 1} + \sum_{n \mathop = 1}^N \alpha_n f_n$

From our induction hypothesis, we have:

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

From the construction of $\sequence {f_n}_{n \mathop \in \N}$ we have:

$f_{N + 1}$ is $\Sigma$-measurable.

So from Pointwise Scalar Multiple of Measurable Function is Measurable, we have:

$\alpha_{N + 1} f_{N + 1}$ is $\Sigma$-measurable.

So, from Pointwise Sum of Measurable Functions is Measurable, we have:

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

$\blacksquare$