Questions about Riemann integrability 
Let $f_n$ be a sequence of Riemann integrable functions that converge uniformly to $f$ on the compact interval $[a,b]$.  Show that $f$ is bounded and Riemann integrable.

I know that bounded functions on compact intervals are exactly those which are Riemann integrable.  I am stuck on showing that the limit function is bounded.
 A: You need to use the uniformity condition.  Remember that $f_n$ converges uniformly to $f$ if for every $\varepsilon$, there is an $N$ such that if $n > N$, then $|f_n(x) - f(x)| <\varepsilon$ for all $x \in [a,b]$. So the $f_n(x)$ converge to $f(x)$ at the same speed, in some sense.  A common way people visualize this is by taking a band of width $2\varepsilon$ around graph of the limit function $f$, i.e. the set $\{(x,y): y \in [f(x)-\varepsilon, f(x)+\varepsilon]\}$.  Now if you graph your $f_n$ too, the uniformity condition implies that $f_n$ lies in this band, if $n$ is sufficiently large.  But if $f$ isn't bounded, that requires your $f_n$ to be unbounded.
To formalize this, the argument goes something like this.  Let $\varepsilon > 0$.  Suppose that $f$ is unbounded.  Then for every $M$, there is an $x$ such that $|f(x)|>M$.  Furthermore, since $f_n$ converges uniformly to $f$, we know that for some $N$, if $n>N$, we have that $|f(x)-f_n(x)|<\varepsilon$.  In particular, if $|f(x)|>M$, $|f_n(x)|>M-\varepsilon$.  Since such an $x$ exists for all $M$, $f_n$ is unbounded.  Contradicts boundedness of $f_n$.  You need to do some work to fix the fact that $f_n$ might be unbounded on some discrete subset of $[a,b]$, but that's the idea.
A: Hint: Use the definition of uniform convergence of a sequence of functions.
