# Does convergence in the sup norm imply convergence in distribution?

I got two sequences of stochastic process $$(X_n(t))_{t \in [0,1]}$$ and $$(Y_{n}(t))_{t \in [0,1]}$$, defined on a probability space $$(\Omega, \mathcal{F},P)$$, and know that their distance in the sup-norm on $$[0,1]$$ converges to $$0$$ almost surely, i.e.

$$\sup \limits_{t \in [0,1]} \vert X_n(t) - Y_n(t) \vert \to 0 \quad P-a.s., \quad n \to \infty$$,

or equivalently

$$P \left (\lim \limits_{n \to \infty} \sup \limits_{t \in [0,1]} \vert X_n(t) - Y_n(t) \vert = 0 \right ) = 1$$.

Now I'm wondering if this also implies the (pointwise) convergence in distribution of $$X_n$$ to $$Y_n$$. This result seems very intuitive, but how does one formally show this?

Thanks!

What you want is Slutsky's theorem. If, for some $$t$$, the sequence $$X_n(t)$$ converges in distribution, and if $$Y_n(t)-X_n(t)$$ converges to $$0$$ in probability, then $$Y_n(t)$$ converges to the same limit law as $$X_n(t)$$. In your case you have almost sure convergence of $$Y_n(t)-X_n(t)$$ to $$0$$, which is stronger than what you need.
• I'm sorry but I don't understand the answer. Do I have convergence in the $r$-th mean in my case above? – Bazzan May 28 at 10:30