Can anyone please check my proof?
We are asked to show that if $f:\mathbb{R}^d \to \mathbb{R}_+$ is a measurable function then $$\int f \ge a \times m(\{f>a\}).$$
I note that $$\int f = \int_{\mathbb{R}_+} m(\{f>t\}) dt \ge \int_{[0,a]} m(\{f>t\}) dt \ge \int_{[0,a]} m(\{f>a\}dt= a \times m(\{f>a\})$$
My first inequality is due to the fact that we have non-negative measure and the second one uses the fact that $\{f>a\} \subset \{f>t\}$ for $t \in [0,a]$ hence $m(\{f>t\}) \ge m(\{f>a\})$.
This is Markov's inequality and is widely used in Probability Theory. See here for a proof.