Is Boundedness a Necessity for Double Integral Proofs?

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engin
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Show that if f is defined on a rectangle R and double integral of f on R
exists, then f is necessarily bounded on R.
 
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If the function is defined on the rectangle the I can't see how it could be unbounded there. Do you mean almost everywhere defined? In the second case, the statement is wrong. Try

[tex]\int_0^1\int_0^1 \frac{y}{\sqrt{x}}dxdy[/tex]
 
engin said:
Show that if f is defined on a rectangle R and double integral of f on R
exists, then f is necessarily bounded on R.

This will be true if you define the integral in terms of ordinary Riemann or Darboux definition. There are more general definitions where boundedness is not required.