Show that $\mu_{f+g}(\lambda_1+\lambda_2) \le \mu_f(\lambda_1)+\mu_g(\lambda_2)$ for $\lambda_1, \lambda_2 \ge 0$

Question

Definition: Distribution Function $\mu_f(\lambda)=\mu\{x \in \Omega : |f(x)| > \lambda\}$

Answer 1

We go to prove that $\{x\in\Omega\mid|(f+g)(x)|>\lambda_{1}+\lambda_{2}\}\subseteq\{x\mid|f(x)|>\lambda_{1}\}\cup\{x\mid|g(x)|>\lambda_{2}\}$ by contradiction. Suppose not, then there exists $x\in LHS\setminus RHS$. $x\notin RHS$ implies that $|f(x)|\leq\lambda_{1}$ and $|g(x)|\leq\lambda_{2}.$ By triangular inequality, $|(f+g)(x)|\leq|f(x)|+|g(x)|\leq\lambda_{1}+\lambda_{2}$, which contradicts to the fact that $x\in LHS$.

Therefore $\mu(LHS)\leq\mu(RHS)\leq\mu\left(\{x\mid|f(x)|>\lambda_{1}\}\right)+\mu\left(\{x\mid|g(x)|>\lambda_{2}\}\right)$. That is, $\mu_{f+g}(\lambda_{1}+\lambda_{2})\leq\mu_{f}(\lambda_{1})+\mu_{g}(\lambda_{2})$.