Let X and Y be random variables with joint pdf $f(x,y)=x+y$ for $0I am using the fact that $X$ and $Y$ are independent if and only if $f_X(x)f_Y(y)=f(x,y)$. So I have
$$f_{X}
(x)=\int_{0}^{1}f(x,y)dy\\=\int_{0}^{1}x+ydy\\=[xy+\frac{y^{2}}{2}]_{y=0}^{y=1}\\=x+\frac{1}{2}$$
and by basically exactly the same math, $f_Y(y)=y+\frac{1}{2}$. Then $$f_X(x)f_y(y)=(x+\frac{1}{2})(y+\frac{1}{2})=xy+\frac{1}{2}(x+y)+\frac{1}{4}\ne f(x,y)$$
And hence they are not independent. But can that be right? Why would the value of X have anything to do with the value of Y? It's not like one is a function of the other. 
Or have I made a simple mistake? I looked through a couple of times and I'm pretty sure my math is right...
 A: Note that the conditional density f(x|y) depends on y.  f(x|y)=f(x,y)/f(y).  
You showed that f(y)=y+1/2 and f(x,y)=x+y.  So f(x|y)=(x+y)/(y+1/2) 
for any 0<=x<=1 and 0<=y<=1.
It clearly is not independent of y. So knowing y does addect the probability that X is in an fixed interval about x.
A: No mistake here. These variables are not independent.
A: X and Y are independent as you have shown. Remember that we are talking about statistical independence. A necessary and sufficient condition for statistical independence is that the joint cumulative distribution function factors as $F_{X,Y}(x,y) = F_X(x) F_Y(y)$. If the joint PDF exists, then an equivalent condition is what you have stated. 
In this problem, we can still pick values of $X$ and $Y$ independently to evaluate the joint PDF or any other function of these two random variables. But $X$ and $Y$ are still statistically dependent by definition.
As a general rule: Statistical dependence $\not\implies$ functional dependence, and functional dependence $\not\implies$ statistical dependence. A famous example of the latter is the case of sample mean and variance of $N$ independent, identically distributed (IID) Gaussian random variables . The sample variance is functionally dependent on the sample mean, but they are statistically independent.
A: Your right. You can also check $E[XY]=E[X]E[Y]$ for the independence condition. This may be clearer because expected values are constant.
Also, independence is interpreted as that even if you know the outcome $X=x$, you cannot use that to guess $Y$: $f(Y|X)=f(Y)$.
For $f(x,y)=x+y$, if you know $X=x$, you have a better idea of what $Y$ will be.
For example, if $X=0.1$ or $X=0.5$, the probability of $Y=0.5$ under these two conditions are different.
