For a merely decreasing positive sequence it fails: $$f_n:=\frac{1}{n}\chi_{[n,\infty)}:\quad\int f_n\mathrm{d}\lambda=\infty\nrightarrow0$$
For a dominated decreasing positive sequence it holds: $$0\leq f_n\downarrow f:\quad\int f_n\mathrm{d}\mu\to \int f\mathrm{d}\mu\quad(f_n\in\mathcal{L})$$ How to prove this via Beppo-Levi?
Since my comment seems to answer the question, let me repeat the argument as an answer...
Denote for shorthand: $\bar{f}:=\sup_n f_n$
By measurability for positive functions one has:
$$\int f_n \mathrm{d}\mu = \int \bar{f} \mathrm{d}\mu + \int (\bar{f}-f_n) \mathrm{d}\mu$$ $$\int f \mathrm{d}\mu = \int \bar{f} \mathrm{d}\mu + \int (\bar{f}-f) \mathrm{d}\mu$$
By monotone convergence one has:
$$\int(\bar{f}-f_n)\mathrm{d}\mu\to\int(\bar{f}-f)\mathrm{d}\mu$$
It remains to observe that all integrals are finite to conclude: $$\int f_n\mathrm{d}\mu=\int\bar{f}\mathrm{d}\mu-\int(\bar{f}-f_n)\mathrm{d}\mu\to\int\bar{f}\mathrm{d}\mu-\int(\bar{f}-f)\mathrm{d}\mu=\int f\mathrm{d}\mu$$