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Do you have any other function that serves as a counterexample to this statement?

If $f$ is Riemann integrable on $ [a, b]$, then $f$ is continuous on $[a, b]$.

My counter-examples

Consider the following
begin{align}f(x)=cases{0& if $x=0,$\sin(1/x)& if $x in (0,1]$.} end{align}

and
begin{align}f(x)=cases{0& if $0leq x<1/2,$\1 & if $1/2leq x leq 1 $.} end{align}
Both are not continuous, but are integrable over $[0,1].$

The following function is also an interesting example-
$G:[0,1]rightarrowmathbb{R}$
$G(x):=begin{cases}
frac{1}{n} for x=frac{1}{n}(ninmathbb{N})\
0 xin[0,1]
end{cases}$

This function is not continuous at infinitely many points($x=frac{1}{n}(ninmathbb{N})$ to be precise) but still turns out to be integrable. Let's prove this by the familiar $epsilon-delta$ definition of Riemann integrability-

Let me guess that the integral is $0$ and then I'll prove it. Consider the set $E={x|G(x)geqfrac{epsilon}{2}}$ which is a finite set(eg-$epsilon=frac{1}{10},E={1,frac{1}{2},frac{1}{3},frac{1}{4},frac{1}{5},frac{1}{6},frac{1}{7},frac{1}{8},frac{1}{9},frac{1}{10}}$ of cardinality n(say) and consider $delta=frac{epsilon}{4n}$ and break the tagged partition $dot{P}$ into two parts

$dot{P_1}$:=tags in $E$ and $dot{P_2}$:=tags in $E^c$ respectively. Then even in the maximal case when all points in $E$ are tags of 2 partitions(i.e are boundary points of both), you would have $S(h;dot{P_1})=2n.frac{epsilon}{4n}$. Also, size of intervals$<1$ which would mean that $S(h;dot{P_2})=frac{epsilon}{2}*(size of interval)<frac{epsilon}{2}$

$S(h;dot{P})=S(h;dot{P_1})+S(h;dot{P_2})< 2n.frac{epsilon}{4n} +frac{epsilon}{2}=epsilon$

Since $epsilon>0$ was any random number, the function is Riemann integrable.

The Thomae function as pointed out in the comments is also an interesting example where the function is not continuous at infinitely many points($mathbb{Q}cap[0,1]$ to be precise) but still turns out to be integrable. For the part about continuity, you can refer to my answer here- https://math.stackexchange.com/a/3056570/584828

For showing that the function is continuous, there are 2 ways that I know of-

1) Using the Lebegue integrability criteria of integrability of a function which says that a bounded function $f:[a,b]rightarrowmathbb{R}$ is Riemann integrable iff it is continuous almost everywhere(meaning discontinuous only at some null set). Clearly, the points of discontinuity are $mathbb{Q}cap[0,1]$(refer to the link above for further explanation) which is a null set and so the function is Riemann integrable.

2) You can also prove it using the familiar $epsilon-delta$ definition of Riemann integrability(this is almost exactly the same as done for G above)-

Let me guess that the integral is $0$ and then I'll prove it. Let $E={x|h(x)geqfrac{epsilon}{2}}$(where h is the Thomae's function) and break the tagged partition $dot{P}$ into two parts

$dot{P_1}$:=tags in $E$ and $dot{P_2}$:=tags in $E^c$ respectively. Now, you must realize that E is a finite set(eg-$frac{epsilon}{2}=frac{1}{5},E={1,frac{1}{2},frac{1}{3},frac{2}{3},frac{1}{4},frac{3}{4},frac{1}{5},frac{2}{5},frac{3}{5},frac{4}{5}}$) of cardinality n(say) and hence, if you set $delta=frac{epsilon}{4n}$, then even in the maximal case when all points in $E$ are tags of 2 partitions(i.e are boundary points of both), you would have $S(h;dot{P_1})=2n.frac{epsilon}{4n}$. Also, size of intervals$<1$ which would mean that $S(h;dot{P_2})=frac{epsilon}{2}*(size of interval)<frac{epsilon}{2}$

$S(h;dot{P})=S(h;dot{P_1})+S(h;dot{P_2})< 2n.frac{epsilon}{4n} +frac{epsilon}{2}=epsilon$

Since $epsilon>0$ was any random number, the function is Riemann integrable.