We have that (in the context of Lebesgue integration)$$\lim_{n \rightarrow \infty} \int_{-\infty}^n f(x) \ dx = 1$$ I wish to show that this implies $\int_{-\infty}^\infty f(x) \ dx = 1$. Is this true by definition of the expression on the left, or is this true by monotone convergence: $$1 = \lim_{n \rightarrow \infty} \int 1_{(-\infty,n]}f(x) \ dx = \int \lim_{n \rightarrow \infty} 1_{(-\infty,n]}f(x) \ dx = \int_{-\infty}^{\infty} f(x) \ dx$$ where MC was invoked at the second equation sign?
EDIT: In above, $f$ is a positive, measurable function.