Originally Posted by

**Moo** I realise that in the steps that would follow what I did, there would be some problems because of absolute values... I don't know if it is possible to solve them, but it makes the proof less simple

There is not really a problem ; for the lower bound, you can write

$\displaystyle P(X_n+Y_n\leq x)\geq P(X_n\leq x-\delta,|Y_n|\leq\delta)$ $\displaystyle =P(X_n\leq x-\delta)-P(X_n\leq x-\delta, |Y_n|>\delta)\geq P(X_n\leq x-\delta)-P(|Y_n|>\delta).$

Then for both the lower and the upper bound, let $\displaystyle \varepsilon>0$ and use the fact that $\displaystyle F_X$ is continuous at $\displaystyle x$ (we prove convergence only at those points $\displaystyle x$) to choose $\displaystyle \delta$ such that $\displaystyle |F_X(x)-F_X(x\pm\delta)|\leq\varepsilon$. In addition, for $\displaystyle n$ large enough, $\displaystyle F_{X_n}(x\pm\delta)$ is close from $\displaystyle F_X(x\pm\delta)$ (it requires that $\displaystyle F_X$ is continuous at $\displaystyle x\pm\delta$ to say that, so we should choose $\displaystyle \delta$ accordingly; there only countably many bad choices so it's OK), and $\displaystyle P(|Y_n|>\delta)$ is small. That should do.