Existence and finiteness of Lebesgue integral for: $f(x)=x^{-1}(e^{-x}-e^{-1/x})$ I think I am getting a little better at these MCT, DCT-type exercises. The issue is to show/prove the existence and finiteness (if they apply) to the following function:
$$f(x)=x^{-1}(e^{-x}-e^{-1/x})$$
Where applicable I want to show as rigorously as possible the justification for existence, etc. For example, it is not enough to say simply, "this exists by MCT"; rather, I need to show that the conditions of measurability, monotone-increasing, etc. are met.
Please comment, add answers, etc. I am learning this on my own so any and all improvements are welcome.
 A: (Note: The way I tried to show measurability in an earlier version of this answer was wrong.)
Since $f$ is continuous, except at $0$, the only possibly issue with measurability is the behavior of $f$ at $x=0$, where $f$ is not defined. The usual 'extension' of the definition is that a function $g$ defined on a measurable set $E \subset X$ is measurable on $X$ iff $\mu E^c = 0$, and $g^{-1}V$ (which is a subset of $E$) is measurable for all open $V$.
So in our case, we take $E = \mathbb{R}\setminus \{0\}$, clearly $m \{0\} = 0$, and if $V$ is open, then $f^{-1}V$ is open by continuity of $f$ on $E$. Hence $f$ is (Borel) measurable.
For sufficiently small $x<0$, and for some $k>0$, you have $|f(x)|>\frac{k}{|x|}$, so $f$ is not integrable. More explicitly, it is straightforward to show that for some $\delta>0$, there exists a $k>0$ such that $|f(x)|>\frac{k}{|x|}$ for $x\in[-\delta,0)$. Then you have (with $0 < \epsilon < \delta$)
$$\int |f| \geq \int_{[-\delta,-\epsilon]} |f| \geq \int_{\epsilon}^{\delta} \frac{k}{x} \, dx \geq \ln \frac{\delta}{\epsilon}.$$
Letting $\epsilon \to 0$ shows that the left hand side is unbounded, hence $f$ is not integrable.
