# When can you integrate a derivative?

Let us say I have an expression $$\frac{d}{dt}f = g$$ where the derivative is taken in a weak sense (of distributions). Can I integrate this from $0$ to $t_0$ and get $$f(t_0) - f(0) = \int_0^{t_0}g?$$

Does the derivative need to be a classical one to do that?

On the face of it, $g$ would have to be a function, and preferably in $L^1$ (or at least, in $L^1_{\mathrm{loc}}$). The condition for this turns out to be that $f$ be absolutely continuous. (Edited to add: Absolute continuity here means that for any $\varepsilon>0$ there is some $\delta>0$ so that, for any finite collection of non-overlapping intervals $[a_k,b_k]$ with $\sum_k|b_k-a_k|<\delta$, we have $\sum_k|f(b_k)-f(a_k)|<\varepsilon$.)
With a suitable interpretation, $g$ could be a signed measure, and $f$ a function of bounded variation. You need to be careful about the ends of the interval at points where $f$ has a jump discontinuity (or $g$ has a point mass), but these points are only countably many.
• Thanks. So you say if $g \in L^1$ this should work. In the wiki, absolute continuity condition is written in an integral form. Do you have a citation for why the integral cancels out the derivative on the LHS though? – matt.w May 9 '13 at 19:48
• Just to clarify: if $g \in L^1(0,t_0)$ then the weak derivative $f'$ is absolutely continuous, so the integration carries through. Thanks, I'll have a look at Royden – matt.w May 9 '13 at 20:05