Show that integrals are equal

Question

Let $f \colon [0,+\infty) \to \mathbb R$ be a convex function. Then $f''(x)$ is a nonnegative distribution on $(0,+\infty)$ and hence it can be continued to a nonnegative mesure $\mu$ on $(0,+\infty)$. How to show that for any continuous function $\varphi(x)$ with compact support on $(0,+\infty)$ we will have $$ \int\limits_0^{+\infty} \varphi(x) \, d\mu(x) = \int\limits_0^{+\infty} \varphi(x) \, df'(x) \; ? $$ For $f \in C^2(0,+\infty)$ it is obviuous. Maybe I should find a sequence of $C^2$-functions $f_n(x)$ such that $f_n(x) \to f(x)$ pointwise, $f_n'(x) \to f'(x)$ in points of continuity of $f'(x)$, $f''(x) \to f''(x)$ in topology of $D'(0,+\infty)$, and pass to the limit?