Real-valued measurable functions $f_n \geq f_{n+1}$ with $f_n \to 0$ pointwise, then $f_n \to 0$ in measure 
Let $(X, \mathcal{A}, \mu)$ be a measure space with $\mu$ finite measure. Let $f_n : (X, \mathcal{A}) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$ be measurable for every $n \in \mathbb{Z}_+$ such that $f_n \to 0$ as $n \to \infty$ pointwise and $f_n \geq f_{n+1}$ for every $n$ pointwise. Does this imply that $f_n \to 0$ in measure?

I think this is false, however, I cannot find a counterexample. The reason I think this is false is that the functions $f_n$ may not be non-negative, therefore we cannot, for instance, let $g_n:= -f_n$ and apply the Monotone Convergence Theorem and the Markov inequality to conclude.

This is true. Suppose that $f_n$ does not converge to $0$ in measure. Then there is an $\varepsilon>0$, $\delta > 0$ and a strictly increasing sequence of naturals $n_k$ such that 
$$\mu(|f_{n_k}| > \varepsilon) > \delta$$ for every $k$.
To conclude, let $$A = \bigcap_{k \in \mathbb{N}} \{ |f_{n_k}| > \varepsilon \}.$$
Notice that $\{|f_{n_{k+1}}| > \varepsilon\} \subseteq \{|f_{n_{k}}| > \varepsilon\}$ so that, since $\mu$ is finite we have that
$$\mu(A) = \lim_{k \to \infty} \mu(|f_{n_k}| > \varepsilon) \geq \delta > 0.$$
But then $f_{n_k} \not \to 0$ pointwise on $A$ which gives a contradiction.
