The standard normal CDF is continuous and strictly increasing

The density $\varphi$ in [F1] is positive and continuous by [F2]. Fix $a<b$. By [F3], $\varphi$ attains a minimum $c$ on $[a,b]$; since $\varphi$ is positive at every point, $c>0$. The density formula and additivity over $(-\mathrm{\infty },a]\bigsqcup (a,b]$ give $\mathrm{\Phi }(b)-\mathrm{\Phi }(a)={\int }_{(a,b]}\varphi (x)dx\ge c(b-a)>0$, so $\mathrm{\Phi }$ is strictly increasing.

Fix $x\in \mathbb{R}$. By [F3], $\varphi$ is bounded by some finite $M$ on the compact interval $[x-1,x+1]$. If $\mathrm{\mid }y-x\mathrm{\mid }<1$, then $\mathrm{\mid }\mathrm{\Phi }(y)-\mathrm{\Phi }(x)\mathrm{\mid }\le M\mathrm{\mid }y-x\mathrm{\mid }$ by integrating the density over the interval between $x$ and $y$. Letting $y\to x$ proves continuity at $x$, and $x$ was arbitrary.

The sets $(-\mathrm{\infty },n]$ increase to $\mathbb{R}$, so [F4] and the total mass in [F1] give ${\lim }_{n\to \mathrm{\infty }}\mathrm{\Phi }(n)=1$. The sets $(-\mathrm{\infty },-n]$ decrease to the empty set and have finite measure, so [F5] gives ${\lim }_{n\to \mathrm{\infty }}\mathrm{\Phi }(-n)=0$. Monotonicity from step 1.1 then gives the two stated limits at infinity.