# Uniform convergence and improper integration

## Homework Statement

Suppose g and f_n are defined on [1,+infinity), are Riamann-integrable on [t,T] whenever 1<=t<T<+infinity. |f_n|<=g, f_n->f uniformly on every compact subset of [1,+infinity), and
$$\int^{\infty}_{1} g(x)dx<\infty$$.
Prove that
$$lim_{n->\infty} \int^{\infty}_{1} f_{n}(x)dx =\int^{\infty}_{1} f(x)dx$$

## The Attempt at a Solution

If I let $$h_{n}(u)=\int^{u}_{1} f_{n}(x)dx$$, then $$lim_{n->\infty} h_{n}(u)=\int^{u}_{1} f(x)dx=h(u)$$ for each u in [1,+infinity). it is equivalent to prove that $$lim_{n->\infty}lim_{u->\infty} h_{n}(u)=lim_{u->\infty}lim_{n->\infty} h_n(u)$$. This is true if h_n converges uniformly to h on [1,+infinity). This is where I got stuck. Actually, I'm not sure if h_n indeed converges uniformly...Or, is there any other way to prove it? Any hint? Thanks a lot!

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Well. I think you can prove uniform convergence of $h_n$ directly.
You want to show that $\forall \varepsilon >0: \exist N>0$ such that whenever n>N $|h_n(u)-h(u)|\leq\varepsilon\quad \forall u\in[1,\infty)$, right?
First pick $u_0\in[1,\infty)$ such that $\int_{u_0}^\infty{dxg(x)}\leq \varepsilon/4$. Then pick an N with $|f_n(x)-f(x)|\leq\frac{\varepsilon}{2(u_0-1)}$ for all n>N and $x\in[1,u_0]$ This N should do the job.

You can write for any u
$$|h_n(u)-h(u)|=\left|\int_1^u{dx[f_n(x)-f(x)]}\right|\leq\left|\int_1^{u_0\wedge u}{dx[f_n(x)-f(x)]}\right|+\chi_{\{u_0<u\}}\left|\int_{u_0}^u{dx[f_n(x)-f(x)]}\right|$$

For n>N the first term is clearly bounded by $\varepsilon/2$. The second term is only there if $u>u_0$. In this case you can use $|f_n(x)|<=g(x)$ for all x (which implies $|f(x)|<=g(x)$ so
$$\left|\int_{u_0}^u{dx[f_n(x)-f(x)]}\right|\leq 2\int_{u_0}^u{dxg(x)}\leq 2\int_{u_0}^\infty{dxg(x)}\leq\varepsilon/2$$
where the last inequality holds by the choice of $u_0$.

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This is brilliant, thanks!
u_0 is exactly what I didn't get! Thanks again!:)

This is brilliant, thanks!
Don't be exaggerating 