# Proof function is integrable

1. Apr 19, 2012

### flurrie

Hello everyone,
I am given the following:
D=[0,1]×[0,1] and f(x,y)=$\frac{1}{1+x+y}$ on D.
a) why is f integrable on D?

I know that the function is integrable because it is bounded and has a finite amount of points where it is discontinuous.
That f is bounded follows directly from the given information.
I also know where f is discontinuous, obviously in the point (0,0)
And I can determine the upper and lower bounds for y and x.
by looking at the boundaries: $\frac{1}{1+x+y}$=0
and $\frac{1}{1+x+y}$=1

so for y:
y=$\varphi$$_{1}$(x)=0
y=$\varphi$$_{2}$(x)=-x
and for x:
x=$\varphi$$_{1}$(y)=0
x=$\varphi$$_{1}$(y)=-y

So I thought that this mend that f(x,y) is discontinuous at the points: (0,0),y=-x,x=-y
but I am not sure because by writing x and y in terms of the other variable I can draw some graphs within D and there the function for y=-x is discontinuous at the boundary x=1 and x=0 and for x=-y is this function discontinuous at y=1 and y=0.

So I'm having a bit of a problem with understanding when a function is discontinuous when there are 2 variables. And I am also wondering, if I can find the points where f is discontinuous, then is it enough to just state that f is discontinuous at those points or do I have to give some kind of proof?

Last edited: Apr 19, 2012
2. Apr 19, 2012

### HallsofIvy

Staff Emeritus
You appear to not understand the basic notation:
"And I can determine the upper and lower bounds for y and x.
by looking at the boundaries: $\frac{1}{1+x+y}= 0$
and $\frac{1}{1+x+y}= 1$"

You are given that "D=[0,1]×[0,1]" which means that the boundaries are the lines x= 0, y between 0 and 1, x= 1, y between 0 and 1, y= 0, x between 0 and 1, and y= 1, x between 0 and 1. The fact that the domainis [0, 1]x[0, 1] says nothing about the value of the function.

Also you say "I also know where f is discontinuous, obviously in the point (0,0)" when, in fact, the function is obviously NOT discontinuous there. How did you arrive at that conclusion?

3. Apr 19, 2012

### flurrie

Yes you're right, when I looked at it later on I saw I was wrong. Somehow I got confused with x and y-simple
But when you look at f = $\frac{1}{1+x+y}$ I can't find any point where it is discontinuous within D=[0,1]×[0,1] . it would be discontinuous if x=0,y=-1 or x=-1,y=0 right?
so would that mean that this function has no discontinuous points on the interval D?

Last edited: Apr 19, 2012