Uniform Convergence of integrals If a sequence of functions $f_n$ are uniformly convergent in a given interval $[a,b]$ to a function $f$, are all riemann integrable, then the integral $$\int ^b_af_ndx\rightarrow\int^b_afdx$$ and $f$ is riemann integrable. but is the convergence uniform?
More precisely, is it true that $\forall \epsilon >0, \exists N:\forall n>N$,
$$\left| \int ^b_af_ndx-\int^b_afdx\right|<\epsilon$$
How would I prove it? Or is it obvious?
 A: It doesn't make sense to speak of uniform convergence in this context. The integrals $$I_n := \int_a^b f_n(x) \, dx \qquad \qquad I:=\int_a^b f(x) \, dx$$ are real numbers which do not depend on another parameter (since $a,b$ are fixed). $$I_n = \int_a^b f_n(x) \, dx \to I=\int_a^b f(x) \, dx \qquad (n \to \infty)$$ is (by the definition of convergence of sequences) equivalent to the following statement: $$\forall \varepsilon>0 \, \exists N \in \mathbb{N} \, \forall n \geq N: |I_n - I| \leq \varepsilon \\ \Leftrightarrow \forall \varepsilon>0 \, \exists N \in \mathbb{N} \, \forall n \geq N: \left|\int_a^b f_n(x) \, dx - \int_a^b f(x) \, dx \right| \leq \varepsilon$$
A: This is tautological. By definition, $\int_a^b f_n$ tends to $\int_a^bf$ means the $\epsilon$ condition you are trying to prove. So if you already know the convergence, there is nothing to do. But actually, I suggest you prove that $\int_a^b f_n$ tends to $\int_a^bf$. It will be a good exercise to clarify these notions.
Hint: the integral is linear and
$$
|\int_a^bg(t)dt|\leq \int_a^b|g(t)|dt\leq (b-a)\sup_{t\in[a,b]}|g(t)|=(b-a)\|g\|_\infty.
$$
