I was reading Grafakos book from classical Fourier Analysis (thir edition, 2o14). Let $(X,\Omega,\mu)$ a measure space and let $f:X \to \mathbb{R}$ a measurable function. First, for $\alpha>0$, he defines $d_{f}(\alpha)=\mu(\{x \in X:|f(x)|>\alpha\})$ (the distribution function of $f$). Then, in a proposition (the proof is an excersice), he states that for $\alpha, \beta>0$ and $f,g$ measurable functions, $$d_{f+g}(\alpha+\beta) \leq d_{f}(\alpha)+d_{g}(\beta).$$ I tried to prove this but i just have $$\{|f+g|>\alpha+\beta\} \subset \{|f|+|g|>\alpha+\beta\}.$$ I also tried to separate $$\{|f+g|>\alpha+\beta\} \subset \{|f|>\alpha\}\cup\{|g|>\beta\},$$ but I think that this isn't true (cause maybe $f$ is small (say $|f|<\alpha$) and $g$ very large (say $|g|>\alpha+\beta$)). So, I don't know how to prove this propiety...
Obs: From the same proposition, I also know that if $|g| \leq |f|$ $\mu-$a.e. implies $d_{g} \leq d_{f}$ (this is obvious because if $|g|>\alpha$, then $|f|>\alpha$) and that $d_{cf}(\alpha)=d_{f}(\alpha/|c|)$ for $c \in \mathbb{C} \setminus \{0\}$ (this follows for a chain of equalities of sets).
Let us assume that $f, g\ge 0$. The inclusion $$\tag{1} \{f+g>\alpha+ \beta \}\subset \{f>\alpha\}\cup\{g>\beta\} $$ is true. To see this, consider the basic set theory fact that $$ A\subset B\cup C\quad \iff \quad B^c\cap C^c \subset A^c, $$ where $X^c$ denotes the complement of $X$. So (1) is true if and only if $$ \{f\le \alpha \}\cap \{g\le \beta\}\subset \{f+g\le \alpha+\beta\}, $$ and this is manifestly true.