Decreasing $f_n$ converges in measure to $0$ implies convergence a.e. to $0$

I'm trying to show that if a sequence of decreasing, measurable function $f_n$ converges to $0$ in measure, then it converges pointwise to $0$ a.e.

I know how to prove that if $f_n$ converges to $f$ in measure, there exists a subsequence $\{f_{n_k}\}$ that converges to $f$ a.e. I know there is a condition that it is a decreasing sequence, and it converges to a constant $0$, but how does that change the fact that the whole sequence (not subsequence) of functions converges to the same value?

Let $A$ be the set of full measure on which $f_{n_{k}}$ converges to $f$. Then given $\varepsilon>0$ and $x\in A,$ there is some $K$ such that for all $k\geq K,$ $|f_{n_{k}}(x)-f(x)|=f_{n_{k}}(x)-f(x)<\varepsilon.$ Now given $n\geq n_{K},$ we see that $|f_{n}(x)-f(x)|=f_{n}(x)-f(x)\leq f_{n_{K}}(x)-f(x)<\varepsilon.$ Then $\forall x\in A,$ $\{f_{n}(x)\}$ converges to $f(x),$ which is sufficient.