Prove the following fact (Probability) I'm trying to prove the following fact

$$P[A\cup B] \ge P[A]$$
  This is what I've done so far, but I got stuck 
  $$P[A \cup B] \ge P[A]$$
  $$P[A] + P[B] - P[A \cap B] \ge P[A]$$
  $$P[A \cap B] + P[A \setminus B] + P[B] - P[A \cap B] \ge P[A]$$
  $$P[A \setminus B] + P[B] \ge P[A]$$
  How can I proceed after this? please help?

 A: Since $\;A\cap B\subseteq B\;$ we get $\;P(A\cap B)\le P(B)\;$ and thus $\;P(B)-P(A\cap B)\ge 0\;$ , so
$$P(A\cup B)=P(A)+\overbrace{P(B)-P(A\cap B)}^{\ge0}\ge P(A)$$
A: Your work looks pretty close, but some tweaking is required.
The point of expressing $A\cup B$ into a union of set differences is that they are disjoint, and the probability of a union of disjoint events is the sum of the probabilities of the events.   (IE: The intersection of disjoint events has zero probability measure.)
This leaves you with a sum of probabilities (all addition, no subtractions), and all probabilities are never less than zero.   So the sum must be never less than just one of the terms alone.
$$Y\geqslant 0 \implies X+Y\geqslant X$$
Since you want to prove the probability is never less than $\mathsf P(A)$, then use $A\cup B= A\cup(B\setminus A)$.


 $$\begin{align} \mathsf P(A\cup B) ~=~& \mathsf P\big(A\cup (B\setminus A)\big) \\[1ex] ~=~ & \mathsf P(A)+\mathsf P(B\setminus A) & \text{union of disjoint events} \\[2ex]\therefore~\mathsf P(A\cup B) ~\geq ~ & \mathsf P(A) &\because~ \mathsf P(B\setminus A) \geq 0 \end{align}$$

