I tried to proof a claim and I'm not sure if I did it right. It would be great if someone could have a look at it! First I give a definiton:
Let $h : [0, \infty ) \rightarrow \mathbb{R}$. We define the convolution operator $*$ for the functions $h*m$ and $h*F$ by: $$ (h*m)(t):=\int\limits_{0}^{t} h(t-x)dm(x), \quad (h*F)(t):=\int\limits_{0}^{t} h(t-x)dF(x),$$ whenever these integrals exist.
Let $h : [0, \infty ) \rightarrow \mathbb{R}$. If the integrals $ (h*m)(t)$ and $(h*F)(t)$ exist, then $*$ is commutative: $$h*m=m*h.$$
My proof:
If I have $$ y(t):= \int\limits_{0}^{t} m(t-y)h(y)dy \Rightarrow y(t)'= \int\limits_{0}^{t} m'(t-y)h(y)dy$$ But I know that the convolution is commutative. So $$ y(t)= \int\limits_{0}^{t} h(t-y)m(y)dy \Rightarrow y(t)'= \int\limits_{0}^{t} h'(t-y)m(y)dy$$ So I have: $$\int\limits_{0}^{t} m'(t-y)h(y)dy=\int\limits_{0}^{t} h'(t-y)m(y)dy$$ $$\Rightarrow (h*m)(t)=\int\limits_{0}^{t} h(t-x)dm(x)=\int\limits_{0}^{t} h(t-x)m'(x)dx =\int\limits_{0}^{t} h(y)m'(t-y)dy=\int\limits_{0}^{t} h'(t-y)m(y)dy=\int\limits_{0}^{t} h'(x)m(t-x)dx=\int\limits_{0}^{t} m(t-x)dh(x)=(m*h)(t)$$
Can I do it like this?