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I have read that the Bergman space $A^p(\Omega)$ consist of all the analytic functions $f$ in $\Omega$, such that $$ \left( \int_{\Omega} |f(z)|^p dA \right)^{1/p} < \infty $$ where $dA$ is the area measure. I am very confuse with what area measure means, I haven't took measure theory yet, my intuition says that $$ \int_{\Omega} |f(z)|^p dA = \iint_{\Omega}|f(x+iy)|^p dxdy $$ Is this right ?? If not how can I compute an integral with $dA$? How about now if $dA_w$ is the weighted area measure?

Leo Sera
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    Your intuition is right. See for example http://en.wikipedia.org/wiki/Bergman_space. – Vectornaut Jun 02 '15 at 22:42
  • @Vectornaut Thanks a lot. So if I understand correctly, the integral with the weighted area measure $dA_w$ will be $$ \int_{\Omega}|f(z)|^pdA_w =\frac{1}{\text{Area}(\Omega)} \int_{\Omega}|f(z)|^pdA=\frac{1}{\text{Area}(\Omega)} \iint_{\Omega} |f(x+iy)|^p dxdy \ ?? $$ – Leo Sera Jun 03 '15 at 00:54
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    It looks like Jonas Meyer answered your question, so I'll just refer you to that, but please let me know if there's something left to be said! – Vectornaut Jun 04 '15 at 02:47

1 Answers1

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The short answer is yes, you can think of the integrals as iterated double integrals as in elementary calculus. Whether there is a normalization constant like $\dfrac{1}{\text{Area}(\Omega)}$ will depend on context (and that is not the only kind of "weighted" area measure, so more context is needed if you want to ask about it).

Area measure means Lebesgue measure on $\mathbb R^2$, restricted to measurable subsets of $\Omega$. Fubini's theorem says that if $f:\mathbb R^2\to \mathbb R$ is integrable, then:

  • For almost all (fixed) $y\in \mathbb R$, the function $\mathbb R\to \mathbb R$ defined by $x\mapsto f(x,y)$ is integrable.
  • The function $\mathbb R\to\mathbb R$ defined by $y\mapsto\int_{\mathbb R}f(x,y)\,dx$ is integrable. (This function is defined almost everywhere by the previous bullet point.)
  • $\int_{\mathbb R}\left(\int_{\mathbb R}f(x,y)\,dx\right)\,dy = \int_{\mathbb R^2}f\,dA$, where the iterated integral exists by the previous bullet points.

This is stated for all of $\mathbb R^2$ for simplicity, but with the usual modifications it applies to your $\Omega$.

On the other hand, the generality of Lebesgue integrability is much more than you need. Because you are only working with continuous functions, no measure theory is needed just to define the integrals: They are Riemann integrals, although possibly improper because the function or region may be unbounded. A continuous Lebesgue integrable function is (maybe improperly) Riemann integrable, and the Lebesgue and Riemann integrals coincide.

An advantage of getting used to the measure-theoretic formulation is that there are useful general results like the dominated convergence theorem that you don't get with Riemann integration.

Jonas Meyer
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