I have $$\int_{\mathbb R} f\,d\mu = \lim_{n\to\infty}\int_{[0,n]} f\,d\mu$$
So $$\begin{align}\int_{\mathbb R}f\,d\mu &= \lim_{n\to\infty}\int_{[0,n]}\chi_{[0,\infty)}e^{-x}\\ &=\lim_{n\to\infty}\int\chi_{[0,n]}e^{-x}\,d\mu \\ &= \lim_{n\to\infty}ne^{-x}=\infty. \end{align}$$
Is this correct?
On
No, that is not correct. You are treating $e^{-x}$ as a constant function in the last equality. $\int \chi[0,n]e^{-x}\,d\mu$ is essentially supposed to be the same as the Riemann itnegral: $\int_{0}^n e^{-x}\,dx$.
The key is that $x$ is the variable of integration. For example, $\chi_{[0,n]}$ should really be written $\chi_{[0,n]}(x)$.
On
Can't post a comment so I will give you this as an answer (sorry). https://en.wikipedia.org/wiki/Dominated_convergence_theorem
No. $$\int \chi[0,n] e^{-x} d\mu(x) \not = n e^{-x}$$ What even is $x$ on the right-hand side?