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## 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

[tex]\int^{\infty}_{1} g(x)dx<\infty[/tex].

Prove that

[tex]lim_{n->\infty} \int^{\infty}_{1} f_{n}(x)dx =\int^{\infty}_{1} f(x)dx[/tex]

## Homework Equations

## The Attempt at a Solution

If I let [tex]h_{n}(u)=\int^{u}_{1} f_{n}(x)dx[/tex], then [tex]lim_{n->\infty} h_{n}(u)=\int^{u}_{1} f(x)dx=h(u)[/tex] for each u in [1,+infinity). it is equivalent to prove that [tex]lim_{n->\infty}lim_{u->\infty} h_{n}(u)=lim_{u->\infty}lim_{n->\infty} h_n(u)[/tex]. 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!