Deciding whether a function is Lebsgue measurable and Lebesgue integrable

Question

Cosnider the function $f:[0, 2]\to \mathbb{R}$, $f(x)=\begin{cases} e^x,&x\in [1, 2]\cup\{0\}\\ n^2,&x\in \left(\frac{1}{(n+1)^2}, \frac{1}{n^2}\right] \text{for } n\in \mathbb{N}. \end{cases}$
a)Decide whether $f$ is Lebesgue measurable.
b)Decide whether $f$ is Lebesgue integrable.
I solved this as follows:
a) The points of discontinuity of $f$ form a countably infinite subset, so the Lebesgue measure of this set is $0$. Thus, $f$ is Lebesgue measurable.
b) The question basically boils down to whether $f$ is Lebesgue integrable on $(0, 1]$. Since the function $\displaystyle A\mapsto \int_Afd\lambda$ is a measure on the Lebesgue measurable subsets of $[0, 2]$, we may write $$\int_{(0, 1]}fd\lambda=\int_{\cup_{n=1}^{\infty}\left(\frac{1}{(n+1)^2}, \frac{1}{n^2}\right]}fd\lambda=\sum_{n=1}^\infty\int_{\left(\frac{1}{(n+1)^2}, \frac{1}{n^2}\right]}fd\lambda=\sum_{n=1}^{\infty}\int_{\left[\frac{1}{(n+1)^2}, \frac{1}{n^2}\right]}fd\lambda=\sum_{n=1}^{\infty}\int_{\frac{1}{(n+1)^2}}^{\frac{1}{n^2}}n^2dx=\sum_{n=1}^{\infty}\frac{2n+1}{(n+1)^2}=\infty.$$ Thus, $f$ is not Lebesgue integrable on $[0, 2]$ because it is not Lebesgue integrable on $(0, 1]$.
I would like to know if my solution is correct because I have just started learning the Lebesgue integral and I am still unsure if I do things correctly.