Measurability and a integral

Question

I need to calculate

$\lim_{n\rightarrow\infty}\int^{\infty}_{0}\frac{cos(\frac{x}{n})}{2^x}d\lambda(x)$

and show that the integral makes sense for every $n$.

My approach so far:

Let $u_n(x):=1_{[0,\infty)}(x)\frac{cos(\frac{x}{n})}{2^x}$. $u_n$ is Borel measurable since $[0,\infty)$ is borel(?) and the fraction is continuous(?)

Since $|1_{[0,\infty)}(x)\frac{cos(\frac{x}{n})}{2^x}|\le 1_{[0,\infty)}(x)\frac{1}{2^x}\in\mathcal{L}^1_{\mathbb{R}}$ for all $n$, we have $u_n\in\mathcal{L}^1_{\mathbb{R}}$ for all $n$

Now by dominating convergence theorem we have

$\lim_{n\rightarrow\infty}\int^{\infty}_{0}\frac{cos(\frac{x}{n})}{2^x}d\lambda(x)=\int^{\infty}_{0}\lim_{n\rightarrow\infty}\frac{cos(\frac{x}{n})}{2^x}d\lambda(x)=\int^{\infty}_{0}\frac{1}{2^x}d\lambda(x)=\frac{1}{ln(2)}$

I'm a little unsure about the measurability of a function, and what $\lambda(x)$ fully implies. Moreover, is this a function that go from any measurable space X into R? what's the distinction between lebesgue measurable and borel measurable?

Also, is [0,infinity) a borel set?

Edit: Fixed some obvious mistakes.

Answer 1

Here the measure space $X$ is the real line endowed with the Lebesgue measure.

The set $[0,+\infty)$ is Borel measurable as the complement of the open set $(-\infty,0)$. a continuous function is indeed Borel-measurable.