Are Riemann integrable functions on a closed and bounded interval continuous? I know continuous function and monotone functions are Riemann integrable, but I’m not sure if Riemann integrable functions are continuous? 
 A: No, try to show that the function (which is clearly not continuous) $$ f(x) = \begin{cases} 0,& x \neq 0,\\ 1,& x=1 \end{cases} $$ is Riemann integrable on $[-1,1]$.
A: No. A bounded function $f\colon[a,b]\longrightarrow\mathbb R$ is Riemann-integrable if and only if it is continuous almost everywhere. But not necessarily everywhere; take for instance$$\begin{array}{rccc}f\colon&[a,b]&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}1&\text{ if }x=a\\0&\text{ otherwise.}\end{cases}\end{array}$$
A: As others have already pointed out, a Riemann integrable function needn't be continuous. Yet, there is another result which is really useful : any Riemann integrable function is bounded (see here for more details If a function $f(x)$ is Riemann integrable on $[a,b]$, is $f(x)$ bounded on $[a,b]$?).
The converse is not true, just take $f:[0,1]\to \mathbb{R}$, $f(x) = \begin{cases} 0,& x \in \mathbb{Q},\\ 1,& x\in \mathbb{R}\setminus \mathbb{Q} \end{cases}$ (a classic application of the Lebesgue Criterion you weere told about).
A: No, not at all. Here is the precise relation of continuity and Riemann-integrability:

Let $f: [a,b] \to \mathbb{R}$ be a bounded function.
Then $f$ is Riemann-integrable if and only if the set of points at
  which $f$ is discontinuous has Lebesgue-measure $0$.

In particular, since countable sets have measure zero, any function that is discontinuous on a countable set and continuous elsewhere will be Riemann-integrable.
So for example the function
$$ \operatorname{sgn}(x): [-1,1] \to \mathbb{R}$$
will be Riemann-integrable, but this is not continuous at $0$.
