If $G(x)=P[X\geq x]$ then $X\geq c$ is equivalent to $G(X)\leq G(c)$ $P$-almost surely 
Suppose $[\Omega,\mathcal{F},P]$ denotes a probability triplet and $X:\Omega\to\mathbb{R}$ is a real-valued random variable. Define
  $$
G(x)=P[X\geq x].
$$
  Claim: for any constant $c$, the event $X\geq c$  is equivalent to $G(X)\leq G(c)$ possibly except for a set of measure $0$. Can you please help me understand this?

Edit: I think I have one direction. Let
$$
A=\{w\in\Omega:X(w)\geq c\},\quad B=\{w:G(X(w))\leq G(c)\}.
$$
Fix $\bar{w}\in A$. I claim that $\bar{w}\in B$. Indeed, $\bar{w}\in A$ implies $X(\bar{w})\geq c$ so that $\{X\geq X(\bar{w})\}\subset\{X\geq c\}$. This gives
$$
P[X\geq X(\bar{w})]\leq P[X\geq c]\iff G(X(\bar{w}))\leq G(c)\iff\bar{w}\in B.
$$
 A: $G(x)=1-P(X<x)\implies G(c)=1-P(X<c)$ 
Now, $X\geq c \implies X|X\geq c := X'=c+\epsilon, \epsilon\geq 0$ where $\epsilon$ is a random variable on the same sample space as $X$
The definition of $X'$ implies:
$G(X')=G(c+\epsilon)=1-[P(X<c)+P(X\in[c,c+\epsilon))]=G(c)-P(X\in[c,c+\epsilon))$
Now, $P(X\in[c,c+\epsilon))$ is a random variable, since $\epsilon$ is a random variable.
By the definition of probability $P(X\in[c,c+\epsilon))\geq0 \implies G(X')=G(c+\epsilon)\leq G(c),\; a.s.$
The other direction
Assume we have some event $A\in \sigma(X):P[a\in A:G(X(a))> G(c)]=0$ (i.e., $A$ is the event that $G(X)\leq G(c), a.s.)$ and let $Y \sim P_X$, then we get:
$P[a\in A: G(X(a))> G(c)]=P[a\in A:P(Y<X(a))<P(Y<c)]=0 \implies A=\{X\geq c,\;a.s.\}$
$\square$
A: Define the event $\mathcal{H}$ according to my hint above (where $\mathcal{H}$ stands for "hint"): 
$$ \mathcal{H} =  \{\omega'\in\Omega: G(X(\omega') ) \leq G(c)\}\cap \{\omega'\in\Omega : X(\omega')< c\} $$
We want to show that $Pr[\mathcal{H}]=0$. 
Your reply to my hint shows you already got most of the way:  I think you agree that: 
$$ \mathcal{H} \subseteq \{\omega'\in\Omega: X(\omega') < c\}\cap \{\omega'\in\Omega : G(X(\omega'))=G(c)\} $$ 
Let $\mathcal{F}$ be the set of all $x \in \mathbb{R}$ such that there is a $y\in\mathbb{R}$ for which $x<y$ and $G(x)=G(y)$ (where $\mathcal{F}$ stands for "flat").  Thus: 
$$\mathcal{H}  \subseteq \{\omega'\in\Omega : X(\omega') \in \mathcal{F}\} $$
and so: 
\begin{align} 
Pr[\mathcal{H}] &\leq Pr[\{\omega'\in\Omega : X(\omega') \in \mathcal{F}\}] \\
&\equiv Pr[X \in \mathcal{F}] 
\end{align}
So it suffices to prove that $Pr[X \in \mathcal{F}] =0$. 

Proving $Pr[X \in \mathcal{F}]=0$. Since the function $G(x)$ is non-increasing, the set $\mathcal{F}$ is either empty, or consists of a finite or countably infinite number of disjoint intervals $\{I_i\}$.  You can show that $Pr[X \in I_i] =0$ for all $i$, and so $Pr[X \in \mathcal{F}]=\sum_iPr[X \in I_i] = 0$. 
A: Since $G$ is decreasing $G(a)\le G(b)$ is equivalent to $b\le a$ except for the constant segments of $G$. 
$a$ and $b$ can be any reals or any real valued functions of any variables, that is even random variables can be plugged in. 
If we plug in a random variable (for, say $a$) whose cdf happens to be  $G$ then the   equivalence will hold a.s.   because                                                                     $X$ falls in a constant segment of $G$  with probability $0$.
