Proving Riemann Integrability- Uniform Convergence Ok so I have come across a proof to show if we have a sequence of functions $f_n$ converging uniformly to $f$ say in the reals, such that if $f_n$ is riemann integrable then so is $f$. In the proof I've come aross there are two "obvious" inequalities that I can't seem to derive which are:
In an interval $I$, and $\epsilon >0$
$$ {\rm sup}\ (f) \leq  {\rm sup}\ (f_n) + \epsilon $$
$${\rm inf}\ (f)\geq {\rm inf}\ (f_n) - \epsilon$$
I know these are somehow derived from the fact $f_n$ converges to $f$ uniformly but I can't get this inequality algebraically, nor does it seem so obvious to me when drawing this out. 
 A: For $\epsilon > 0$, there is $N \in \mathbb{N}$ such that
$$
|f(x) - f_n(x)| < \epsilon \quad\forall n\geq N, x \in I
$$
Whence
$$
f(x) < f_N(x) + \epsilon \leq \sup_{n\in \mathbb{N}} f_n(x) + \epsilon \quad\forall x\in I
$$
and so
$$
f < \sup f_n + \epsilon
$$
A: Alternatively we can use the Riemann-Lebesgue Theorem to show this along with some facts regarding spaces of functions. 
Suppose the sequence of functions $(f_n)$ is Riemann Integrable, then this sequence is bounded and we know that there exists a zero set $Z_n$ such that $(f_n)$ is continuous at every point except those in $Z_n$. Given that if $(f_n)$ is a sequence of continuous functions, and if each $f_n$ is continuous at $x_0$ we know that $f$ is continuous at $x_0$, we can conclude that $f$ is continuous at every point except the union of the zero sets of each $f_n$. But the countable union of zero sets is also a zero set, and hence it is a zero set for $f$. Since $f$ is bounded and its set of discontinuity points forms a zero set, the Riemann-Lebesgue theorem tells us that $f$ is Riemann Integrable. 
