Statement of the problem: Show that if $\{f_n\}_{n=1}^{\infty}$ is a sequence of non-negative measurable functions on $\mathbb{R}^d$, then for any $t>0$ we have: $$m\left(\left\{x\in \mathbb{R}^d:\enspace \liminf_{n\to\infty}f_n(x)>t\right\} \right) \leq \liminf_{n\to\infty}m\left(\left\{x\in \mathbb{R}^d:\enspace f_n(x)>t\right\} \right).$$
Attempt/Discussion When I first looked at the statement I immediately thought of Fatou's Lemma which in this case would tell us that (directly following the statement of the Lemma): $$\int_{\mathbb{R}^d}\liminf_{n\to\infty} f_n dm\leq \liminf_{n\to\infty}\int_{\mathbb{R}^d}f_n dm.$$ We then have the relations (which can be proved from Fubini's Theorem): $$ \int_{\mathbb{R}^d}\liminf_{n\to\infty} f_n dm=\int_0^\infty m\left(\left\{x\in \mathbb{R}^d:\enspace \liminf_{n\to\infty}f_n(x)>t\right\} \right)dt$$ and $$\int_{\mathbb{R}^d}f_n dm=\int_0^\infty m\left(\left\{x\in \mathbb{R}^d:\enspace f_n(x)>t\right\} \right)dt $$ This then implies $$\int_0^\infty m\left(\left\{x\in \mathbb{R}^d:\enspace \liminf_{n\to\infty}f_n(x)>t\right\} \right)dt \leq \liminf_{n\to\infty}\int_0^\infty m\left(\left\{x\in \mathbb{R}^d:\enspace f_n(x)>t\right\} \right)dt.$$
Although this statement looks close, it is not the result of the statement. It seems that the statement in the problem Implies Fatou's lemma but not necessarily the other way around (I'm not sure yet). If anyone has any thoughts or ideas they would be appreciated!
Note that $\liminf f_n(x) > t$ if and only if $f_n(x) > t$ for all but at most finitely many indices $n$. Thus $$\left\{x : \liminf f_n(x) > t \right\} = \bigcup_{n=1}^\infty \bigcap_{k \ge n} \{x : f_k(x) > t\}.$$ Use continuity from below: $$m \left( \bigcup_{n=1}^\infty \bigcap_{k \ge n} \{x : f_k(x) > t\} \right) = \lim_{n \to \infty} m \left( \bigcap _{k \ge n} \{x : f_k(x) > t\} \right).$$ Since $$\bigcap_{k \ge n} \{f_k(x) > t\} \subset \{f_n(x) > t\}$$ you obtain $$ \lim_{n \to \infty} m \left( \bigcap _{k \ge n} \{x : f_k(x) > t\} \right) = \liminf_{n \to \infty} m \left( \bigcap _{k \ge n} \{x : f_k(x) > t\} \right) \le \liminf_{n \to \infty} m(\{x : f_n(x) > t\}).$$