Prove that $e^{-x}I(x \geq 0)$ is a measurable function in the Lebesgue measure space given by $(R, \mathcal{B}, \lambda)$

Question

How would I prove this? I want to use the definition that a function $X: \Omega \rightarrow R$ ($X: R \rightarrow R$ in this case) is measurable if for $x \in R$, the set $[\omega \in \Omega: X(\omega) \leq x]$ is measurable (belongs in $\mathcal{B}$). Would this be the right method, and if so, how do I move forward with this?

Answer 1

Hint. You know that $I(x \geq 0)$ is measurable. Then, all you have to do is prove that $e^{-x}$ is measurable, since the product of measurable functions is measurable too.

Simpler. Notice that since $f:= x \mapsto e^{-x}$ is a continuous function, we have $f^{-1}((-\infty, x])$ is closed and belongs to $\mathcal{B}$.

Proof using that it is a limit of simple functions. For that, you can write $e^{-x}$ as an increasing limit of simple functions, and use that the limit of measurable functions is measurable. Try to make the usual construction of partioning into dyadic intervals; this approach is similar to the construction of integral of a positive measurable function with respect to a measure done in Bartle's book.