Show $\sum_{k=1}^\infty \frac{(1)^k}{k}1_{[k,k+1)}$ is Borel measurable

Question

I have the following function $u=\sum_{k=1}^\infty \frac{(1)^k}{k}1_{[k,k+1)}$

I want to show that this is Borel measurable. My idea is to show $\{u \geq a\} \in \mathcal{B}(\mathbb{R})$ according to my textbook. For a=1/2 I get $\emptyset \in \mathcal{B}(\mathbb{R})$. And for a=-1 I get $\mathbb{R} \in \mathcal{B}(\mathbb{R})$

But I don't how to show mathematically for other values of a.

Any hint would be appreciated

Answer 1

There are some basic theorems about measurability. Sums of measurable functions are measurable and pointwise limits of measurable functions are measurable . Since each term in your sum is measurable the result follow form these theorems.

Answer 2

For $a \leq -1$ you have the same conclusion as in the case of $a=-1$

So we have to examine all $a \in (-1,+\infty)$

Take an $a \in (-1,+\infty)$.

Then since the intervals $[k,k+1)$ are disjoint,then $\{x:u(x)>a\}$ is going to be either the empty set, or a union of intervals for the form $[m,m+1)$ with the interval $(-\infty,0)$, or just a union of intervals of the form $[m,m+1)$ where $m \in \Bbb{N}$

So in every case $\{u>a\}$ is a Borel set.