Question on convolution in measure theory to probability theory and the need for independence In a classical measure theory context, with two measures $\mu_{1},\; \mu_{2}$ on $\mathcal{B}(\mathbb R)$ as the image of the product measure $\mu_{1}\otimes \mu_{2}$ under the map $+:\mathbb R\times \mathbb R \to \mathbb R$ is defined as a convolution. When switching over to probability theory, I notice that we also introduce the notion of independence, i.e. for independent random variables $X$ and $Y$ the convolution is the distribution of the sum $X+Y$.
Question: Why do we need the notion of independence for the random variables? Surely by the definition of the convolution in measure theory it immediately follows that we are evaluating the sum of the random variables? Or am I missing something obvious? I guess the notion of not knowing what the product measure looks like could be the pitfall. Thanks in advance!
 A: Remember that for any probability measure $\mu$ on $\mathcal B(\mathbb R)$ and measurable map $f:\mathbb R\to\mathbb R$, if $\mu$ is the distribution of a random variable $Z$, then the image of $\mu$ by $f$ is the distribution of $f(Z)$.
Here if $\mu_1\otimes\mu_2$ is the distribution of $Z$, then the convolution $\mu_1\star\mu_2$ is the distribution of $+(Z)$. Denoting by $X$ and $Y$ the marginals of $Z$, that is the random variables such that $Z=(X,Y)$, we have that if $(X,Y)\sim\mu_1\otimes\mu_2$, then $X+Y\sim\mu_1\star\mu_2$. But saying that $(X,Y)\sim\mu_1\otimes\mu_2$ is equivalent to say that $X$ and $Y$ are independent (it can even be taken as a definition of independence).
Indeed, you probably saw that $X$ and $Y$ are independent iff for any measurable Borel sets $A,B\in\mathcal B(\mathbb R)$, $\mathbb P(X\in A,Y\in B)=\mathbb P(X\in A)\mathbb P(Y\in B)$. Denoting by $\mu_1$ the distribution of $X$, $\mu_2$ the distribution of $Y$ and $\mu$ the distribution of $(X,Y)$, the latter equality can be rewritten as
$$
\forall A,B\in\mathcal B(\mathbb R),\quad\mu(A\times B)=\mu_1(A)\mu_2(B),
$$
which means that $\mu=\mu_1\otimes\mu_2$.
