Let $X$, $Y$ be independent random variables. Assuming that $X$ is continuous ($\forall_{x\in\mathbb{R}}P(X=x)=0)$. Show that $X+Y$ is continuous.
My approach:
Let $x\in\mathbb{R}$. \begin{align} P(X+Y=x)&=P(\bigcup_{y\in\mathbb{R}}\{X=y \land Y=x-y\})\\ &\leq\sum\limits_{y\in\mathbb{R}}P(X=y\land Y=x-y)\\ &=\sum\limits_{y\in\mathbb{R}}P(X=y)P(Y=x-y) \end{align}
And since $(\forall_{x\in\mathbb{R}}P(X=x)=0)$ we get:
$$\sum\limits_{y\in\mathbb{R}}P(X=y)P(Y=x-y)=\sum\limits_{y\in\mathbb{R}}0\cdot P(Y=x-y)=0$$
Is this proof correct?
Your argument works only when $X$ is discrete. Measures do not behave well with uncountable unions. The correct argument uses Fubini;s Theorem: $P\{X+Y=z\}=\int P\{X=z-y\}dF_Y(y)=0$ because $P\{X=z-y\}=0$ for all $y$.