TITLE: Prove that $a \in \mathbb{R} (P(X + Y = a) \le \gamma)$ if $\forall a \in \mathbb{R} \left(P(X = a) \le \gamma \wedge P(Y=a) \le \gamma\right)$
QUESTION [1 upvotes]: Let $X$ and $Y$ be independent random variables such that $P(X = a) \le \gamma$ and $P(Y=a) \le \gamma$ for each $a \in \mathbb{R}$. Prove that $P(X + Y = a) \le \gamma$ for every $a \in \mathbb{R}$. Is independence of the variables a necessity?

REPLY [1 votes]: It is enough if one of the random variables satisfies the condition. 
If e.g. $P(X=a)\leq\gamma$ for every $a\in\mathbb R$ then for a fixed $a\in\mathbb R$:
$$\begin{aligned}P\left(X+Y=a\right) & =\int1_{\left\{ a\right\} }\left(x+y\right)dF_{X,Y}\left(x,y\right)\\
 & =\int\int1_{\left\{ a\right\} }\left(x+y\right)dF_{X}\left(x\right)dF_{Y}\left(y\right)\\
 & =\int P\left(X=a-y\right)dF_{Y}\left(y\right)\\
 & \leq\int\gamma dF_{Y}\left(y\right)=\gamma
\end{aligned}
$$
Here the second equality is a consequence of independence.
If there is no independence then things go wrong if e.g. $Y=a-X$ for some constant $a$.

edit (alternative solution that avoids integrals)
Let $I:=\{i\in\mathbb R\mid P(X=i)>0\text{ or }P(Y=i)>0\}$.
Then $I$ is countable and consequently for a fixed $a\in\mathbb R$ the sets $J_a:=\{\langle i,j\rangle\in I^2\mid i+j=a\}$ and $K_a:=\{j\in I\mid a-j\in I\}$ are countable.
Then:
$$\begin{aligned}P(X+Y=a) & =\sum_{\langle i,j\rangle\in J_{a}}P\left(X=i,Y=j\right)\\
 & =\sum_{\langle i,j\rangle\in J_{a}}P\left(X=i\right)P\left(Y=j\right)\\
 & \leq\sum_{\langle i,j\rangle\in J_{a}}\gamma P\left(Y=j\right)\\
 & =\sum_{j\in K_{a}}\gamma P\left(Y=j\right)\\
 & =\gamma\sum_{j\in K_{a}}P\left(Y=j\right)\\
 & \leq\gamma\cdot1=\gamma
\end{aligned}
$$
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