[proofplan] Independence makes the law of $(X,Y)$ equal to $\mu\otimes\nu$. The law of $X+Y$ is the pushforward of that product measure under addition, which is the convolution measure. [/proofplan] [step:Compute the pushforward] Let $A\in\mathcal B(\mathbb R)$. Since $X$ and $Y$ are independent, the law of $(X,Y)$ is $\mu\otimes\nu$. Hence \begin{align*} \mathbb P(X+Y\in A) &=(\mu\otimes\nu)(\{(x,y):x+y\in A\}) \\ &=\int_{\mathbb R}\int_{\mathbb R}1_A(x+y)\,d\nu(y)\,d\mu(x). \end{align*} [/step] [step:Identify the convolution] By definition of convolution of probability measures, \begin{align*} (\mu*\nu)(A)=\int_{\mathbb R}\int_{\mathbb R}1_A(x+y)\,d\nu(y)\,d\mu(x). \end{align*} Thus $\mathbb P(X+Y\in A)=(\mu*\nu)(A)$ for every Borel set $A$, so the law of $X+Y$ is $\mu*\nu$. [/step]