Let $X$ be an absolutely continuous random variable, i.e., the cumulative distribution function is given by
$$F(x)=\int_{-\infty}^xf(t)\,dt$$
for some nonnegative (integrable) function $f$. Then it is easy to show using calculus that $F$ is continuous.
Now, let $X$ and $Y$ be random variables such that the joint cumulative distribution function is given by
$$F(x,y)=\int_{-\infty}^x\int_{-\infty}^yf(s,t)\,ds\,dt$$
for some nonnegative (integrable) function $f(x,y)$. Then does it follow that $F$ is continuous?
Best Answer
It certainly does. For any integrable function $f$ on any measure space $\int_A |f|d\mu \to 0$ as $\mu (A) \to 0$. Continuity of your function follows easily from this.
You can also prove this using DCT and sequential definition of continuity. Suppose $x_n \to x$ and $y_n \to y$. Then $\int 1_{\{(t,s):t \leq x_n, s\leq y_n\}} f(t,s)dtst \to \int 1_{\{(t,s):t \leq x, s\leq y\}} f(t,s)dtst $ because the integrand is dominated by $|f(t,s)|$ which is integrable and the intergand converges a.e. to $f(t,s)$. Note that $\iint |f(t,s)|dtst=(\int |f(t)|dt)^{2}<\infty$ by Tonelli's Theorem.