problem on uniform convergence of a riemann integrable function

Let $f_n$ and $f$ be continuous functions on an interval [$a, b$] and assume that $f_n → f$ uniformly on $[a, b]$. Pick out the true statements:
(a) If $f_n$ are all Riemann integrable, then $f$ is Riemann integrable.
(b) If $f_n$ are all continuously differentiable, then $f$ is continuously differentiable.
(c) If $x_n → x$ in [a, b], then $f_n(x_n) → f(x)$.

how should i able to solve this problem .somebody help me please.

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If this is a homework problem, please tag it as "homework". –  Ron Gordon Dec 31 '12 at 12:28
What have you tried? Which statements do you think are true, and why? –  Asaf Karagila Dec 31 '12 at 12:49
i am not sure how to start.give any clue please –  gomti Dec 31 '12 at 12:53

Hints:

(a) Try proving $|\overline{\int^b_a} f(x) dx-\underline\int_a^b f(x)\,dx|\le 2\epsilon_n(b-a).$ Where $\epsilon_n=$sup$|f_n(x)-f(x)| ,$ keep in mind $\epsilon_n \to0$ as $n\to \infty$

(b) If $f'_n(x)$ is continuous,and if it converges uniformly use $\displaystyle\lim_{x\to\infty}\int^b_af'(x)dx=\int^b_ag(x)$ and show $g(x)=f'(x)$

(c) is straightforward

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Hints:

(a) A function is Riemann integrable if

$\quad(i)$ it is bounded, and

$\quad(ii)$ has a countable set of discontinuities.

(c) you need the fact that if $g$ is a continuous, then
$$x_n \to x \implies g(x_n)\to g(x)\quad \mathbb{as}\quad n\to \infty,$$
$$|f_n(x_n)-f(x)|=|(f_n(x_n)-f_n(x))+(f_n(x)-f(x))|$$
$$\leq |f_n(x_n)-f_n(x)|+|f_n(x)-f(x)| < \dots.$$