Given a function $f:[a,b]\to \mathbb R$ it could make sense, depending on the situation, to define the integral of $f$ in either of the following ways:
- Standard Riemann integral, assuming it exists.
- Limit as $\varepsilon\to 0$ of the Riemann integral over $[a+\varepsilon, b]$, assuming the latter integral exists for sufficiently small $\varepsilon$.
Does anyone know an example of a function $f$ for which these two do not coincide? For instance a function for which the second method is well-defined but the Riemann integral does not exist?
Consider the function $f(x)= n$ for $x=1/n$ for $n$ a natural number and $f(x)=0$ otherwise.
For any $\varepsilon > 0$ on the interval $[\varepsilon, 1]$ this function is bounded and zero up to finitely many points. Hence the Riemann integral exists and is zero. So using definition 2. the limit $\varepsilon \rightarrow 0$ exists and is equal to $0$.
However, on the interval $[0,1]$ this function has infinitely many discontinuities, is unbounded and is not Riemann integrable according to definition 1.