Let $X$ be a random variable and $P(X<u)\leq \Phi (u)\leq P(X\leq u)$ for all $u$. Everything is defined on the real line and $\Phi(\cdot)$ is standard Normal.
Is X a continuous random variable and $F_X(u)=\Phi(u)$?
My intuition tells me that this is true. Graphically, $F_X$ is always right continuous and if $\Phi(u)\neq P(X\leq u)$ then there would be a contradiction by increasing $u$ by an epsilon amount.
(Follow-up questions, edited material)
How would you prove it?
My attempt:
$$\lim_{n\rightarrow \infty} \Phi(u-\frac{1}{n})\leq \lim_{n\rightarrow \infty} P(X<u)\leq \lim_{n\rightarrow \infty} \Phi (u)\leq\lim_{n\rightarrow \infty} P(X\leq u)\leq \lim_{n\rightarrow \infty} \Phi(u+\frac{1}{n})$$
Thus,
$$\Phi(u)\leq P(X<u)\leq\Phi(u) \leq P(X\leq u)\leq\Phi(u)\ \blacksquare$$
Can you guys provide an alternative way to prove it?