Is $\lim_{n\to\infty}\int_0^1f_n(x)=\int_0^1f(x)$ in this case?

Question

Let $f_n(x)=\begin{cases}\frac{e^{x^2}}{x^2} &x\in\left[1/n,1\right]\\ 0 &x\in(-\infty,1/n)\cup(1,+\infty)\end{cases}.$

This converges to $f(x)=\begin{cases}\frac{e^{x^2}}{x^2} &x\in\left(0,1\right]\\ 0 &x\in(-\infty,0]\cup(1,+\infty)\end{cases}$ pointwise on $\Bbb R$ and uniformly on $(-\infty,-\varepsilon]\cup[\varepsilon,+\infty)$ for every $\varepsilon>0$.

I'm then asked to find $\lim_{n\to\infty}\int_0^1f_n(x)dx$, and I think the following holds: $$\lim_{n\to\infty}\int_0^1f_n(x)dx=\lim_{n\to\infty}\int_0^{1/n}0+\lim_{n\to\infty}\int_{1/n}^1\frac{e^{x^2}}{x^2}dx=\lim_{n\to\infty}\int_{1/n}^1\frac{e^{x^2}}{x^2}=\int_{0}^1\frac{e^{x^2}}{x^2}=+\infty.$$Am I correct?

Answer 1

You are correct. Also, you can use Dominated convergence Theorem with the Dominant function $f$ observing that $|f_n(x)|\leq |f(x)|$ $\forall n$ $\forall x$ since $$\frac{e^{x^2}}{x^2}\geq 0 \hspace{2mm}\forall x\in (0,1].$$ Thus, by the Dominated convergence theorem,

$$\lim_{n\rightarrow \infty}\int f_n=\int \lim_{n\rightarrow \infty}f_n =\int f =\int_{(0,1]}\frac{e^{x^2}}{x^2}.$$

Since $$ +\infty=\int_{(0,1]}\frac{1}{x^2}\leq \int_{(0,1]}\frac{e^{x^2}}{x^2}, $$

we have $$\lim_{n\rightarrow \infty}\int f_n=\int_{(0,1]}\frac{e^{x^2}}{x^2}=+\infty.$$

Answer 2

Yes it is true. This follows from Beppo Levi's monotone convergence theorem for Lebesgue integral here: https://en.wikipedia.org/wiki/Monotone_convergence_theorem

Answer 3

$f_n(x) \ge \frac{1}{x^2}$ on $[1/n,1]$ and by positivity of Riemann integral $$ \int_0^1 f_n(x) dx = \int_{1/n}^1 \frac{e^{x^2}}{x^2} dx \ge \int_{1/n}^1 \frac{1}{x^2} dx = n-1.$$

The result follows.