How to prove $f(x)=e^{\frac{1}{x}}$ is continous in $(0,a), a>0 $ and $\int_{0}^{a}e^{\frac{y}{x}}dx, y>0$ does not exist

Question

I would aprecciate any advice. I'm trying to prove that in the context of a measure space, $(X,B,\lambda)$ , with $X=(0, + \infty) $, $B$ the Borel sigma-algebra and $\lambda$ the Lebesgue measure, there exist some continous function $f : (0,a) \mapsto \mathbb{R}$ such that $f \notin L_p(\lambda)$ for all $p \in (0, \infty)$, there was a hint about using $f(x)=e^{\frac{1}{x}}$.

Answer 1

By a change of variables $u=\frac{y}{x}$, $$\int_0^a e^{\frac{p}{x}} d\lambda(x)=y\int_{y/a}^\infty \frac{e^u}{u^2} d\lambda(u)$$

But $e^{u}/u^2>\frac{1}{u^2}+\frac{1}{u}+\frac{1}{2}$ is not integrable at $\infty$.