Have you ever seen the construction for Volterra's function? Is is a function that is everywhere differentiable, its derivative is bounded everywhere, but this derivative is not Riemann-integrable. You might be able to use a similar construction to find a function that is continuous at zero, but not Riemann-integrable on any interval containing zero. You just need the function to be discontinuous on a set of positive measure in every open interval containing zero, yet still continuous at zero.
Anyway, you said that
was finite, but you did not say it was bounded. So, I actually need to revise the "sketch" for the proof. I was using the assumption that it was bounded. Since it is continuous at zero, given any
, there exists
such that for all
,
. Hence,
for any
. This follows from the definition of continuity.
So, let's choose
such that for all
, we have
. The Riemann integral is supposed to give the area under a curve. That area is bounded by the minimum and maximum height of the function. Since we know that
will never go below
, that is the lower bound for the value of the function. We know it will never go above
. This is true on any subinterval of
. I will replace the
that you are using with
so as not to confuse it with the
we used for continuity to find
. So, for any
, the rectangle with base from
to
has a length of
. It has a width (or height) of at minimum
and at maximum
. So,
(since the integral in the middle is a sum of areas of rectangles, and we know none of those rectangles can have heights outside of that range). Since
is finite and
is finite, we have:
Then, by the Squeeze Theorem, we have
.
Again, all of this depends on
being Riemann-integrable on some open interval containing zero.