Suppose $f\geq 0$, and $f$ is lebesgue integrable. If $\alpha>0$ and $E_\alpha=\{x:f(x)>\alpha\}$, prove that $$ m(E_\alpha)\leq 1/\alpha \int f $$
My proof attempt
Proof. Let the assumptions be as above. Then $$ \alpha 1_{E_\alpha}<f(x)1_{E_\alpha} $$ Then by monotonicity of the integral $$ \alpha\int 1_{E_\alpha}<\int f(x)1_{E_\alpha}\leq \int f(x) $$ $\implies m(E_\alpha)<1/\alpha\int f$.
My main concern is that I end up with $<$ instead of $\leq.$ Any feedback on the proof or style is much appreciated.
Firstly you have not specified what $x$ is, and note that $\alpha \mathbf1_{E_\alpha}<f(x)\mathbf1_{E_\alpha}$ is not true for all $x$. What we have is that $\alpha\mathbf1_{E_\alpha}(x)<f(x)\mathbf1_{E_\alpha}(x)$ for all $x\in E_\alpha$. For all $x\notin E_\alpha$ we have $\alpha\mathbf1_{E_\alpha}(x)=0=f(x)\mathbf1_{E_\alpha}(x)$. Thus what we have in general is that $a\mathbf1_{E_\alpha}\leq f\mathbf1_{E_\alpha}$. From this point on your proof seems fine.