Just a longer comment - I'll show that improper integrals are continuous.
Improper integral on $(0,1]$ is just the limit of regular integrals on $[\delta,1]$. If the limit $L$ exists, then for $\delta < \delta_0$ this integral is $\varepsilon$-close to L. From the triangle inequality we know, that two integrals on $[\delta,1]$ and $[\delta',1]$ are $2\varepsilon$-close (note that $\delta,\delta'$ might also be zero here, then we get just $\varepsilon$). It means that when we integrate on an interval included in $[0,\delta_0]$, then the result is smaller than $2\varepsilon$. Moreover on $[\delta_0,1]$ the function is bounded by some $M$ (from its integrability). So, if we take an interval $[a,b]$ shorter than $\min(\delta_0,\varepsilon/M)$, then (by triangle inequality) the integral is smaller than $2\varepsilon + \varepsilon=3\varepsilon$, which shows the continuity of an improper integral.