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 Homework Statement:

Let ##(f_n)## be a sequence of integrable functions on ##[a,b]##, and suppose ##f_n \rightarrow f## uniformly on ##[a,b]##. Prove that ##f## is integrable and
$$\int_a^b f = \lim_{n\rightarrow\infty} \int_a^b f_n$$
 Relevant Equations:
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##\textbf{Attempt at solution}##: If I can show that ##f## is integrable on ##[a,b]##, then for the second part I get :
Let ##\frac{\varepsilon}{ba} > 0##. By definition of uniform convergence, there exists ##N = N(\varepsilon) > 0## such that for all ##x \in [a,b]## we have ##\vert f(x)  f_n(x) \vert < \frac{\varepsilon}{ba}##. This gives us,
$$\vert \int_a^b f(x)dx  \int_a^b f_n(x)dx \vert = \vert \int_a^b f(x)  f_n(x) dx \vert \le \int_a^b \vert f(x)  f_n(x) \vert dx < \int_a^b \frac{\varepsilon}{ba}dx = \varepsilon$$ when ##n > N##.
It follows ##\lim_{n\rightarrow\infty} \int_a^b f_n(x) dx = \int_a^b f(x) dx## for all ##x \in [a,b]##. []
For the first part I am stuck. Let ##\varepsilon > 0##. Then I need a partition ##P## of ##[a,b]## such that ##U(f, P)  L(f, P) < \varepsilon##. We're given that for any ##n##, there is a partition ##P## such that ##U(f_n, P)  L(f_n, P) < \varepsilon##. How can I proceed?
Let ##\frac{\varepsilon}{ba} > 0##. By definition of uniform convergence, there exists ##N = N(\varepsilon) > 0## such that for all ##x \in [a,b]## we have ##\vert f(x)  f_n(x) \vert < \frac{\varepsilon}{ba}##. This gives us,
$$\vert \int_a^b f(x)dx  \int_a^b f_n(x)dx \vert = \vert \int_a^b f(x)  f_n(x) dx \vert \le \int_a^b \vert f(x)  f_n(x) \vert dx < \int_a^b \frac{\varepsilon}{ba}dx = \varepsilon$$ when ##n > N##.
It follows ##\lim_{n\rightarrow\infty} \int_a^b f_n(x) dx = \int_a^b f(x) dx## for all ##x \in [a,b]##. []
For the first part I am stuck. Let ##\varepsilon > 0##. Then I need a partition ##P## of ##[a,b]## such that ##U(f, P)  L(f, P) < \varepsilon##. We're given that for any ##n##, there is a partition ##P## such that ##U(f_n, P)  L(f_n, P) < \varepsilon##. How can I proceed?