Proving the sum rule of probability limits

Question

I'm trying to show that if $X_n$ converges in probability to 0 and $Y_n$ converges in probability to 0, then $X_n+Y_n$ converges in probability to $0$, ie the sum rule for probability limits.

What I'm trying to do is to show it directly using $\epsilon, \delta$ arguments as opposed to appealing to the CMT, to improve my understanding of convergence. But I get stuck almost immediately because I don't know how I can use bounds for the probability each sequence exceeds some value to get a bound for the probability the sum exceeds some value...

Here's where I get stuck:

For every $\epsilon>0$ and every $\delta>0$ I have an $N$ s.t $n>N \implies P( |X_n|>\epsilon)<\delta$ and $P(|Y_n|>\epsilon)<\delta$. I also know that $|X_n+Y_n| \leq |X_n|+|Y_n|$ and that therefore $P( |X_n+Y_n|>\epsilon) \leq P( |X_n|+|Y_n| >\epsilon)$. Now it seems obvious that if $P( |X_n|>\epsilon)$ and $P(|Y_n|>\epsilon)$ can be made as small as we want, the RHS can be bounded, but I can't quite work out how. I can't seem to relate the probability of the sum to the probabilities I can bound.

Answer 1

You can argue like this. If $\left| X_n \right| + \left| Y_n \right| > \epsilon$, then it must be the case that $\left| X_n \right| > \frac{\epsilon}{2}$ or $\left| Y_n \right| > \frac{\epsilon}{2}$. Because if both $\left| X_n \right|, \left| Y_n \right| \leq \frac{\epsilon}{2}$, then:

$$ \left| X_n \right| + \left| Y_n \right| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon $$

So now you can use the union bound

$$ Pr( \left| X_n \right| + \left| Y_n \right| > \epsilon ) \leq Pr \left( \left| X_n \right| > \frac{\epsilon}{2} \lor \left| Y_n \right| > \frac{\epsilon}{2} \right) \leq Pr \left( \left| X_n \right| > \frac{\epsilon}{2} \right) + Pr \left( \left| Y_n \right| > \frac{\epsilon}{2} \right) $$

which you can bound in terms of $\delta$.