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Let $(X,\mathcal{F})$ be a measurable space and let $E:\mathcal{F}\to\mathscr{B(H)}$ be a spectral measure. Let $\phi\in B(X)$ be a simple function whose image is $\{\lambda_1,\ldots,\lambda_n\}\subset\mathbb{C}$, define

$\intop_X \phi dE = \sum_{i=1}^n\lambda_i\cdot E(\phi^{-1}(\lambda_i))$

Now, for a general $\phi\in B(X)$ let $\phi_n\to\phi$ uniformly, and define

$\intop_X \phi dE = \lim\intop_X \phi_n dE$

Prove that this integration operator is uniquely defined.

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A limit is missing. Do you know what you have to show? –  Davide Giraudo Jun 9 '12 at 12:42
    
Yeah, I've actually managed to since I've posted this question... –  Shai Deshe Jun 9 '12 at 13:09
    
Well, first show that the limit indeed exist, then that it doesn't depend on the approximating sequence –  Davide Giraudo Jun 9 '12 at 13:12
    
In fact when we are stuck at a math exercise, it can be for two reason: either we don't know what we have to show, or we know it but we don't see what it will be true. I asked you in which of these two cases you were in order to give the best help I can. –  Davide Giraudo Jun 9 '12 at 13:14
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-1: Surely you know by now that asking a homework question calls for more than just copying the problem. –  Nate Eldredge Jun 9 '12 at 13:56

1 Answer 1

First, the sequence $\{\int_X\phi_ndE\}\subset \mathcal B(\mathcal H)$ is Cauchy. Indeed, let $m,n$ integers. Then $$\left\lVert \int_X\phi_mdE-\int_X\phi_ndE\right\rVert\leq \lVert \phi_m-\phi_n\rVert_{\infty}\lVert I\rVert=\lVert \phi_m-\phi_n\rVert_{\infty}.$$ Now, we have to show that the limit doesn't depend on the approximating sequence $\{\phi_n\}$. If $\{\psi\}$ is a sequence of bounded measurable functions on $X$ which converges uniformly to $\phi$ on $X$, then $$\left\lVert \int_X\phi_ndE-\int_X\psi_ndE\right\rVert\leq \lVert \phi_n-\psi_n\rVert\leq \lVert \phi_n-\phi\rVert+\lVert \phi-\psi_n\rVert.$$

(we used $\left\lVert\int_X fdE\right\rVert\int_X \left\lVert f\right\rVert dE$)

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