Re: Integral help?


adomplayer@xxxxxxxxx wrote:
Robert Israel wrote:
In article <1162348018.792066.47050@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<adomplayer@xxxxxxxxx> wrote:
Let f:[0,1]x[0,1] be defined by f(x,y) = (x-1/2)^(-3) if y < |x-1/2|,
f(x,y) = 0 otherwise. What can we say about int_0^1 int_0^1 f(x,y)
dxdy, int_0^1 int_0^1 f(x,y)dydx, and int_E f(x,y) (dy X dx) where
E=[0,1]^2 and (dy X dx) is the product measure?

OK, so what do you get when you do those iterated integrals?

Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

If you do them in one order, the triangle to the left of x=1/2
contributes -infty and the triangle to the right of x=1/2 contributes
+infty, so I *think* it is rigorous to say the total volume then is
undefined?

I didn't ask about the "volume", I asked about the iterated integrals.
Yes, int_0^1 int_0^1 f(x,y) dx dy = int_0^1 |x - 1/2|^(-2) signum(x -
1/2) dx
is a divergent integral.

If you do them in another order, each horizontal slab contributes 0,
making the total volume 0.

Again, it's not a "volume".
Yes, int_0^1 f(x,y) dx = 0 for all y > 0, so int_0^1 int_0^1 f(x,y) dx
dy = 0.

But as for the product measure, I have no idea how to compute product
measures when Tonelli-Fubini fail. I mean, Fubini is basically the
canonical way to compute product measures, to the point where it's very
often used as the *definition* of product measures, in basic calculus
courses. It would be very helpful if you could give me some pointers
how to make sense of/compute the integral of the function wrt the
product measure, for the above function.

By definition, a real-valued measurable function is integrable iff its
absolute value is
integrable. But in this case it's easy to see that the integral of
|f(x,y)| is
infinite.

Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

.

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