Inequality concerning the distribution function 
Definition. Let $(X,\mu)$ be a measure space. For $f$ a measurable function on $X$, the distribution function of $f$ is the function $d_f$ defined on $[0,\infty)$ as follows: $$d_f(\alpha) := \mu\left( \left\{ x \in X : \left| f(x)\right|>\alpha\right\}\right) $$

Now I would like to proove $$d_{f + g}(\alpha + \beta) \leqslant d_f(\alpha) + d_g(\beta)$$
for $\alpha,\beta > 0$. Any suggestions?  
Edit. I think this should work also for $\alpha,\beta \geqslant 0$.
 A: Show and use the fact that for $\alpha$ and $\beta$, 
$$\left\{x\in X,\left|f(x)+g(x)\right|\gt \alpha+\beta\right\}  \subset \left\{x\in X,\left|f(x)\right|\gt \alpha\right\}\cup \left\{x\in X,\left|g(x)\right|\gt \beta\right\}.$$
This can be done by considering the (reversed) inclusion of complements, that is 
$$\left\{x\in X,\left|f(x)\right|\leqslant \alpha\right\}\cap \left\{x\in X,\left|g(x)\right|\leqslant \beta\right\}  \subset \left\{x\in X,\left|f(x)+g(x)\right|\leqslant \alpha+\beta\right\}.$$
A: We have that
$$\left\{ x \in X : \left| f(x)+g(x)\right|>\alpha+\beta\right\}\subseteq
\left\{ x \in X : \left| f(x)\right|>\alpha\right\}\cup \left\{ x \in X : \left| g(x)\right|>\beta\right\}.$$
In fact, if $x$ is in the set on the left and it is not in the set on the right, then $| f(x)|\leq \alpha$, $| g(x)|\leq \beta$, and
$$\alpha+\beta\geq\left| f(x)|+|g(x)\right|\geq \left| f(x)+g(x)\right|>\alpha+\beta$$
which is a contradiction.
Then the inequality follows from the fact that if $C\subseteq A\cup B$, then
$$\mu(C)\subseteq\mu(A\cup B)\leq \mu(A)+\mu(B).$$
