Does the convergence of $\int_{-\infty}^\infty f_n(x) \,dx $ imply the convergence of $\int_{-\infty}^{\infty}f(x)\,dx$?

Question

The sequence of functions, $f_n: \mathbb{R}\to\mathbb{R}$, is bounded and converges uniformly to $f$. Fix any $n \in \mathbb{N}$, we have $\int_{-\infty}^\infty f_n(x)\,dx$ converges. Does the convergence of $\int_{-\infty}^\infty f_n(x)\,dx$ imply the convergence of $\int_{-\infty}^\infty f(x)\,dx$?

I think the answer is yes.

My attempt is to prove the contrapositive statement:

Suppose $\int_{-\infty}^\infty f(x)\,dx$ diverges, define a positive increasing sequence of numbers, $\{a_k\},$ such that $a_k \to \infty$ as $k\to\infty$. Then $\forall M\gt0$, $\exists K \in \mathbb{N}$, s.t. $k\gt K $ implies $\int_{-a_k}^{a_k} f(x) \gt M$, which implies $\int_{-a_k}^{a_k} |f(x)| \, dx \gt M$.

Since $f_n$ converges to $f$ uniformly, given ${\varepsilon\over2a_k}$, $\exists N$ s.t. $n>N$ implies $|f_n-f| \lt {\varepsilon\over2a_k}$.

By triangular inequality, $|f_n|\gt|f|\,-\,{\varepsilon \over 2a_k}$

Taking integrals for both sides, $\int_{-a_k}^{a_k}|f_n(x)| \, dx \gt M - \varepsilon$, implying the divergence of $\int_{-a_k}^{a_k} |f_n(x)| \, dx$.

However, I don't know how to go on and disprove the convergence of $\int_{-a_k}^{a_k}f_n(x) \, dx$

Answer 1

The statement is false. Suppose that$$f_n(x)=\begin{cases}\frac x{1+x^2}&\text{ if }x\in[-n,n]\\0&\text{ otherwise}\end{cases}$$and that $f(x)=\frac x{1+x^2}$. Then $(f_n)_{n\in\Bbb N}$ converges uniformly to $f$, each integral $\int_{-\infty}^\infty f_n(x)\,\mathrm dx$ converges, but $\int_{-\infty}^\infty f(x)\,\mathrm dx$ diverges.

Answer 2

Let $f_n(x) = \max\{n,1/x^2\}.$ Then $\displaystyle\int_{-\infty}^{+\infty} f_n(x) \, dx < +\infty$ and $$ \int_{-\infty}^{+\infty} \lim_{n\to\infty} f_n(x)\, dx = \int_{-\infty}^{+\infty} \frac{dx}{x^2} = +\infty. $$