Show that F(x,y) is not a joint CDF Given $F(x,y)= 1$ if $y\geq e^{-x}$ and is $0$ otherwise.

I understood that $F(x,y)$ equals $1$ in the region above the curve. From this graph, how do I conclude that $F(x,y)$ is not a CDF?
 A: The interpretation of the question by the commentators and in the answer by SmallChess is that it is asking for a proof that $e^{-x}, -\infty < x < \infty$, or $e^{-x}\mathbf 1_{\{x\colon 0 < x < \infty\}}$ is not a valid (one-dimensional) CDF. This interpretation is apparently also shared by the OP who has accepted SmallChess's answer. I think that this interpretation trivializes the problem because it is hard to think of even one property of (one-dimensional) CDFs that these functions do satisfy, and so a proof that they are not (one-dimensional) CDFs is hardly worth asking for on stats.SE.  
A much more interesting problem (certainly one with more meat in it) is the one asked in the title of the question: prove that
$$F(x,y) = \begin{cases} 1, & y \geq e^{-x},\\0, & y < e^{-x},\end{cases}$$
is not a valid joint CDF. Note that $F(x,y)$ seemingly does have
 all the properties that a valid joint CDF must necessarily have: it is a nondecreasing right-continuous function  in each variable, with limit $0$ as the variable tends to $-\infty$ and limit $\leq 1$ as the variable tends to $\infty$. Nonetheless, $F(x,y)$ is not a valid joint CDF because it does not have what I call the rectangle property of (two-dimensional) joint CDFs:

For all $a,b,c,d \in \mathbb R$ such that $a<b$ and $c<d$
$$F(b,d) - F(b,c) - F(a,d) + F(a,c) \geq 0.$$

Note that the expression above is just $P\{a < X \leq b, c < Y \leq d\}$ which of course must be nonnegative.  Applying this to $a=0, b=1, c = \frac 12, d = 1$, we have that
$$F(1,1) - F(1,\frac 12) - F(0,1)+ F(0,\frac 12) = 1-1-1+0 = -1.$$
Thus, the rectangle property does not hold for this function and so it cannot be a valid joint CDF.
A: PS: It's the probability density function that gives area below the curve equal to 1. Not above.
CDF stands for "cumulative distribution function". When your curve is "cumulative" the probabilities, it should not "trend" downward. 
This is the CDF for normality:

Note the curves don't go downward as in your question. Also the curves eventually approach cumulative probability of 1.0. There is no possibility the diagram you have for your question is a CDF.
