All the books I have seen have proved that, for a normal bounded operator $T$, there is a unique spectral measure $E$ such that $\int_{\sigma(T)}^{}\lambda\,dE=T$ by first proving in it for a general Algebra. I am trying to modify this proof to get straight to $\int_{\sigma(T)}^{}\lambda\,dE=T$ without proving it for a general algebra.

I have almost immediately come up against a problem because it uses Riesz representation theorem and I can't think how to integrate (pardon the pun) that into my proof. Surely though there is a way to do this since I am trying to prove something easier. Can anyone help?

Note: I also posted this on stack exchange as I am not sure which is more suitable.

Thanks

All the books I have seen have proved that, for a normal bounded operator $T$, there is a unique spectral measure $E$ such that $\int_{\sigma(T)}^{}\lambda\,dE=T$ by first proving in it for a general Algebra. I am trying to modify this proof to get straight to $\int_{\sigma(T)}^{}\lambda\,dE=T$ without proving it for a general algebra.

I have almost immediately come up against a problem because it uses the Riesz representation theorem and I can't think how to integrate (pardon the pun) that into my proof. Surely though there is a way to do this since I am trying to prove something easier. Can anyone help?

Note: I also posted this on stack exchange as I am not sure which is more suitable.

Thanks

All the books I have seen have proved that, for a normal bounded operator $T$, there is a unique spectral measure E such that $\int_{\sigma(T)}^{}\lambda\,dE=T$ by first proving in it for a general Algebra. I am trying to modify this proof to get straight to $\int_{\sigma(T)}^{}\lambda\,dE=T$ without proving it for a general algebra. I have almost immediately come up against a problem because it uses riesz representation theorem and I can't think how to integrate (pardon the pun) that into my proof. Surely though there is a way to do this since I am trying to prove something easier. Can anyone help?

Note: I also posted this on stack exchange as I am not sure which is more suitable.

Thanks

All the books I have seen have proved that, for a normal bounded operator $T$, there is a unique spectral measure $E$ such that $\int_{\sigma(T)}^{}\lambda\,dE=T$ by first proving in it for a general Algebra. I am trying to modify this proof to get straight to $\int_{\sigma(T)}^{}\lambda\,dE=T$ without proving it for a general algebra. 

I have almost immediately come up against a problem because it uses Riesz representation theorem and I can't think how to integrate (pardon the pun) that into my proof. Surely though there is a way to do this since I am trying to prove something easier. Can anyone help?

Note: I also posted this on stack exchange as I am not sure which is more suitable.

Thanks