Let $\left\{f_n\right\}$ and $f$ be measurable functions from a measure space $(X,\mathcal{M},\mu)$ to a metric space $(Y,d)$ (with the Borel $\sigma$-algebra). Suppose $f_n\rightarrow f$ in measure; that is, for all $\epsilon>0$ and all $\delta>0$ there exists an $N$ such that for all $n\ge N$, $$\mu(x\in X: d(f_n(x), f(x))\ge\epsilon)\le \delta.$$
Is convergence in measure equivalent to the following? For all $\epsilon>0$ there exists an $N$ such that for all $n\ge N$, $$\mu(x\in X: d(f_n(x), f(x))\ge\epsilon))\le \epsilon.$$
I remember seeing a result like this, but cannot remember where. If true, a proof or reference would be very much appreciated. Even an assurance of truth would be great! Want to use this result for a different proof.
Yes they are. Assuming the second one is valid, then for $\epsilon>0$ and $\delta>0$, consider $\eta=\min\{\epsilon,\delta\}$, there is an $N$ such that $\mu(\cdots\geq\eta)\leq\eta$. But now $(\cdots\geq\epsilon)\subseteq(\cdots\geq\eta)$, so $\mu(\cdots\geq\epsilon)\leq\mu(\cdots\geq\eta)\leq\eta\leq\delta$, so we have $\mu(\cdots\geq\epsilon)\leq\delta$.