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**SlipEternal** 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 $\displaystyle \psi(x)$ 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 $\displaystyle \epsilon>0$, there exists $\displaystyle \delta>0$ such that for all $\displaystyle |x| < \delta$, $\displaystyle |\psi(x)-\psi(0)|<\epsilon$. Hence, $\displaystyle \psi(0)-\epsilon < \psi(x) < \psi(0)+\epsilon$ for any $\displaystyle |x|<\delta$. This follows from the definition of continuity.

So, let's choose $\displaystyle \delta>0$ such that for all $\displaystyle -\delta < x < \delta$, we have $\displaystyle \psi(0)-1 < \psi(x) < \psi(0)+1$. 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 $\displaystyle \psi(x)$ will never go below $\displaystyle \psi(0)-1$, that is the lower bound for the value of the function. We know it will never go above $\displaystyle \psi(0)+1$. This is true on any subinterval of $\displaystyle (-\delta,\delta)$. I will replace the $\displaystyle \epsilon$ that you are using with $\displaystyle t$ so as not to confuse it with the $\displaystyle \epsilon=1$ we used for continuity to find $\displaystyle \delta>0$. So, for any $\displaystyle 0<t<\delta$, the rectangle with base from $\displaystyle -t$ to $\displaystyle t$ has a length of $\displaystyle 2t$. It has a width (or height) of at minimum $\displaystyle \psi(0)-1$ and at maximum $\displaystyle \psi(0)+1$. So, $\displaystyle 2t(\psi(0)-1) \le \int_{-t}^t \psi(x)dx \le 2t(\psi(0)+1)$ (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 $\displaystyle \psi(0)-1$ is finite and $\displaystyle \psi(0)+1$ is finite, we have:

$\displaystyle \lim_{t \to 0} 2t(\psi(0)-1) = 0 = \lim_{t\to 0} 2t(\psi(0)+1)$

Then, by the Squeeze Theorem, we have $\displaystyle \lim_{t\to 0} \int_{-t}^t \psi(x)dx = 0$.

Again, all of this depends on $\displaystyle \psi(x)$ being Riemann-integrable on some open interval containing zero.