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Spectral/projection-valued measures have very handy applications theoretically, but I got stuck when asked to compute explicitly certain projection-valued measures. Let's focus on the following:

Let $ N: {\mathcal{L}^{2}}([0,1]) \to {\mathcal{L}^{2}}([0,1]) $ be the normal operator defined by $$ \forall f \in {\mathcal{L}^{2}}([0,1]),\forall t \in [0,1]: \quad [N(f)](t) \stackrel{\text{def}}{=} t \cdot f(t). $$ What is the projection-valued measure corresponding to $ N $?

As $ \sigma(N) = [0,1] $, we need a projection-valued measure $ P $ supported on $ [0,1] $. Theoretically, it should be defined just by $$ P(E) = {\chi_{E}}(N), $$ where $ \chi_{E} $ is the characteristic function of $ E \subseteq [0,1] $ and $ {\chi_{E}}(N) $ is obtained via the Borel functional calculus of $ N $.

However, to find $ {\chi_{E}}(N) $, we need to find a sequence of polynomial functions converging to $ \chi_{E} $, which is not an easy job.

I somehow feel that the projection-valued measure is given simply by $ P(E) = {\chi_{E}}(N) $, but I am not sure.

Can someone give a hint on how to compute the spectral/projection-valued measure explicitly?

Thanks!

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    You can't explicitly compute this projection since you don't have explicit description of Borel sets.2012-12-13
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    @Hui Yu: It is not true that every bounded Borel-measurable function on a complete metric space is the pointwise limit of a sequence of polynomial functions. This is because the pointwise limit of a sequence of continuous functions on a metric space, if such a limit exists, must be continuous on a dense set of points. You can prove this using the Baire Category Theorem.2013-01-07
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    @Hui Yu: Hence, $ \chi_{[0,1] \cap \mathbb{Q}} $ cannot be the pointwise limit of a sequence of polynomial functions. Instead of pointwise convergence, one can ask for pointwise convergence *almost everywhere*. This is possible by Lusin's Theorem.2013-01-07
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    @Norbert: Strange. The Borel subsets are staring at me right in my face. Are they not just the usual Borel subsets of $ [0,1] $?2013-01-07
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    @HaskellCurry Borel sets are of very complicated structure. You can use intersection (finite or countable), complementation, union (finite or countable) and other operations to construct Borel sets. But even after applying them countable amount of times you will not get all the Borel sets. You must make this procedure up to $\omega_1$ oridinal to cover the whole variety of Borel sets. This is what I meant when I said that structure of Borel sets is complicated.2013-01-09

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