I need to show that if $|f_n|\leq g$, and $g$ is integrable, then $\{f_n\}$ is uniformly integrable, i.e., $\underset{a\rightarrow\infty}\lim\sup_n\int_{[|f_n|\geq a]} |f_n| d\mu=0$.
Here is how I thought about it: Since $|f_n|\leq g$, and $g$ is integrable, then each $f_n$ is integrable. Thus, for each $n$, $|f_n| I_{[|f_n|\geq a]}\leq |f_n|$ and $|f_n| I_{[|f_n|\geq a]}\rightarrow 0$ almost everywhere, thus, by Lebesgue dominated convergence theorem, we have $\underset{a\rightarrow\infty}\lim\int_{[|f_n|\geq a]} |f_n|d\mu=0$ for each $n$, and hence, $$0=\sup_n \lim_{a\rightarrow\infty}\int_{[|f_n|\geq a]}|f_n|d\mu=\lim_{a\rightarrow\infty}\sup_n\int_{[|f_n|\geq a]}|f_n|d\mu.$$
However, I am not so sure about the last step of interchanging the limit and supremum? Any thoughts?
It might work to consider that $\sup_n |f_n|$ itself is measurable and bounded by $g$, so it's integrable; and $$\lim_{a \rightarrow \infty} \int_{|f_k| \ge a} \sup_n |f_n| \, \mathrm{d}\mu = 0 \; \; \mathrm{for} \; \mathrm{every} \; k$$ by the argument you gave above. This should be enough.