Can limit go inside the integral if sequence has only one point of non uniform convergence?

Question

Let $\displaystyle f_{n}(x) =\frac{1}{1+n^{2}x^{2}}$ for $n\in \mathbb{N}$, $x\in \mathbb{R}$.Which of the following are true?

1) $f_{n}$ converges pointwise on $[0,1]$ to a continuous function.

2) $f_{n}$converges uniformly on [0,1].

3) $f_{n}$ converges uniformly on $[\frac{1}{2},1]$.

4) $\displaystyle \lim_{n\to\infty}\int_{0}^{1} f_{n}(x) dx=\int_{0}^{1}(\displaystyle \lim_{n\to\infty} f_{n}(x)) dx.$

Using $M_{n}$ test we can prove this series is not uniformly convergent. And point of nonuniform convergence is 0. Hence we can say it will be uniformly convergent on $[\frac{1}{2},1]$ but since it is not uniformly convergent on $\mathbb{R}$ can we say 4th option is not true?

Answer 1

It is very easy to calculate $\int_{0}^{1} f_{n}(x) dx$ and $\int_{0}^{1}(\lim_{n\to\infty} f_{n}(x)) dx$.

Then you can check if option 4 is valid.

Answer 2

Note that in this case $$\int_{0}^{1} f_{n}(x) dx=\int_{0}^{1}\frac{1}{1+n^{2}x^{2}},dx=\frac{1}{n}[\arctan(nx)]_{x=0}^1=\frac{\arctan(n)}{n}.$$ What may we conclude about option 4?

More generally, if the sequence $(|f_n|)_n$ is uniformly bounded by $M>0$ in $(0,1]$ (in your case we can take $M=1$) and $f_n\to 0$ uniformly in $[a,1]$ for any $0<a<1$, then $\int_{0}^1 f_n(x) dx\to 0$. In fact, we have that for $\epsilon>0$, eventually $$\left|\int_{0}^1 f_n(x) dx\right|\leq \int_{0}^{\epsilon/(2M)} |f_n(x)| dx +\int_{\epsilon/(2M)}^1 |f_n(x)| dx\\\leq \frac{\epsilon}{2M}\cdot M+\int_{\epsilon/(2M)}^1 |f_n(x)| dx \leq \frac{\epsilon}{2}+ \frac{\epsilon}{2}= \epsilon$$ where in the final step we used the uniform convergence in the interval $[\epsilon/(2M),1]$.