Neat proof to show that $\sup\{x\; : P(X<x)< \lambda\}=\inf\{x\; : P(X\leq x)\geq \lambda\}$

Question

Consider a probability space $(\Omega,\mathcal{F},P)$ and some real-valued random variable $X$. I want to show that $\sup\{x\; : P(X<x)< \lambda\}=\inf\{x\; : P(X\leq x)\geq \lambda\}$ for any $\lambda \in (0,1)$.

My idea:

By monotonicity of the probability measure: for any $x \in \mathbb R$ we have:

$P(X<x)\leq P(X\leq x)$ and consider that sets

$$ A:=\{x:P(X< x)< \lambda\}\;\;, B:= \{x:P(X\leq x)\geq \lambda\}$$

For any $x^{*}\in A\cap B$, we have:

$$ P(X<x^{*})< \lambda \leq P(X\leq x^{*})$$ such that$P(X=x^{*})>0$ and $x^{*}$ satisfies:

$$\sup\{x\; : P(X<x)< \lambda\}=x^{*}=\inf\{x\; : P(X\leq x)\geq \lambda\} $$

Now assume that $A\cap B= \varnothing$, then $x^{a} \in A$ and $x^{b}\in B$ such that $x^{a}<x^{b}$ such that $\sup A\leq \inf B$. Furthermore, since $A\cap B= \varnothing$, we can say that there exists $x$ such that $P(X\leq x)=\lambda$

I feel like I may be making this over-complicated looking at different cases.

Answer 1

For $x<\sup A$, there exists $x'\in A$ with $x'>x$ and hence $P(X\le x)\le P(X<x')<\lambda$, i.e., $x\notin B$. For $x>\sup A$, there exists $x'<x$ with $x'\notin A$, hence $P(X\le x)\ge P(X<x')\ge \lambda$, i.e., $x\in B$. Thus with $s:=\sup A$, we find that either $B=(s,\infty)$ or $B=[s,\infty)$. At any rate $\inf B=s$.