Basic probability proof I'm currently looking through the proof of this lemma and I am getting stuck at this (trivial!) stage of the proof and I was wondering if someone could help? I've written up the part of the proof that I am stuck on -
Lemma Let X, Y be random variables, let a be a real number and ε > 0. Then 
$P(Y\leq{a})\leq P(X\leq a+\epsilon) + P(|Y-X|>\epsilon)$
Proof
$P(Y\leq{a})=P(Y\leq a, X\leq a +\epsilon) + P(Y\leq a, X> a +\epsilon)$
$\leq P(X\leq a +\epsilon) + P(Y-X\leq a-X, a-X< -\epsilon)$
$\leq P(X\leq a +\epsilon) + P(Y-X< -\epsilon)$
$\leq P(X\leq a +\epsilon) + P(Y-X< -\epsilon) + P(Y-X> \epsilon)$
$ = P(X\leq a+\epsilon) + P(|Y-X|>\epsilon)$
The proof seems to imply that $P(Y\leq a, X\leq a +\epsilon) \leq P(X\leq a +\epsilon)$ 
But I don't understand why this is true?
 A: It's always a good idea to reduce such manipulations to the axioms of probability (and, if needed, simple algebra).
Begin by noting we can decompose the sample space  $\Omega$ into the union of any event and its complement:
$$\Omega = \{|X-Y| \gt \epsilon\}\ \bigcup\ \{|X-Y| \le \epsilon\}.$$
This breaks the event $Y \le a$ into two disjoint parts,
$$\{Y \le a\} = \left(\{Y \le a\} \cap \{|X-Y| \gt \epsilon\}\right)\ \bigcup\ \left(\{Y \le a\} \cap\{|X-Y| \le \epsilon\}\right).\tag{1}$$
Because these parts are disjoint, their probabilities will add (that's an axiom).  Let's tackle the evaluation of the probabilities separately.
Left hand side
Another axiomatic relationship is that the probability of a subset of an event is never greater than the probability of the event itself.  This allows us to overestimate the left hand portion of $(1)$:
$$\Pr\left(\{Y \le a\} \cap \{|X-Y| \gt \epsilon\}\right)\le \Pr(|X-Y| \gt \epsilon).\tag{2}$$
This is intuitively clear.
Right hand side
The right hand portion is due to the algebraic fact that inequalities add:
$$Y \le a\text{ and } |X-Y|\le \epsilon\ \text{ implies }\ X= Y + (X-Y) \le a + \epsilon.$$
(This is readily deduced from the triangle inequality.)
Therefore the right side of $(1)$ is a subset of a simpler event $X \le a+\epsilon$.  Once again this gives an inequality of probabilities
$$\Pr\left(\{Y \le a\} \cap\{|X-Y| \le \epsilon\}\right) \le \Pr\left(X \le a+\epsilon\right).\tag{3}$$
Intuitively, if $Y$ does not exceed $a$ and $X$ is within $\epsilon$ of $Y$, then $X$ cannot exceed $a+\epsilon$.
Solution
Taking probabilities in $(1)$  by means of the results $(2)$ and $(3)$ yields
$$\Pr(Y \le a) \le \Pr(|X-Y| \gt \epsilon) + \Pr\left(X \le a+\epsilon\right),$$
QED.
Putting this result in words might help the intuition:

The chance of $Y \le a$ cannot be greater than the chance that $X \le a+\epsilon$ (if $X$ is within $\epsilon$ of $Y$) plus the chance that $X$ is more than $\epsilon$ away from $Y$.

A: Let the us define the following events
$A = \{X \leq a+e\}$, $B = \{X \leq a+e,Y\leq a\}$, and $C = \{X \leq a+e,Y>a\} $
Now,
\begin{equation}
A = B \bigcup C
\end{equation}
\begin{equation}
\implies B \subseteq A \implies P(B) \le P(A)
\end{equation}
A: That part is easy as you suspect. Let the event A={X<=a+e) and B= {Y<=a} then you have P(A and B) <= P(A) simply because the event  {A and B} is a subset of A.
