Let $g:\mathbb{R} \rightarrow [0,\infty)$, where $g(x)=x^2$. Calculate $\int gd\lambda$, where $\lambda$ is the lebesgue measure.
I've done the following, but I am not sure if this is correct:
$$\int x^2d\lambda =\lim_{n\rightarrow \infty}\int x^21_{[-n,n]}d\lambda=\lim_{n\rightarrow \infty}R\int_{-n}^{n}x^2dx=\lim_{n\rightarrow \infty}\left(\frac{n^3}{3}-\frac{-n^3}{3}\right)=\infty$$
On
You are making things more complicated than necessary, and can avoid MCT and other stuff.
Just let $f:\mathbb R\to[0,\infty)$ be the indicator function of $[1,\infty)$ and observe that $0\leq f(x)\leq g(x)$ for each $x\in \mathbb R$.
That implies directly that: $$\int gd\lambda\geq\int fd\lambda=\lambda([1,\infty))=+\infty$$
If $R$ is used to denote the Riemann-Integral, then what you write is correct:
$$ \begin{align} \int x^2d\lambda &=\lim_{n\rightarrow \infty}\int x^21_{[-n,n]}d\lambda \text{ (Monotone Convergence)} \\ &=\lim_{n\rightarrow \infty}R\int_{-n}^{n}x^2dx \text{ (Riemann=Lebesgue for continuous functions on compact intervals)}\\ &=\lim_{n\rightarrow \infty}\left(\frac{n^3}{3}-\frac{-n^3}{3}\right) \\ &=\infty \end{align} $$