Regularizing measures

  • Thread starter lark
  • Start date
In summary, Rudin's Real and Complex Analysis discusses how a measure can be regularized in order to make it unique. If X is locally compact and Hausdorff, and every open set is \sigma-compact, then \Phi is associated with a regular measure \mu^\prime by \Phi(f) = \int_Xfd\mu^\prime. This means that all the complex measures on X are regular. However, there are examples of measures that are not regular.
  • #1
lark
163
0
Reading Rudin's Real and Complex Analysis, a question in the .pdf attached.
(and no pressure about using the built-in Latex, please)
Laura
 

Attachments

  • quest.pdf
    31.6 KB · Views: 232
Physics news on Phys.org
  • #2
Can you explain how you're "regularizing" [itex]\mu[/itex]?

And I suppose a Borel measure on a [itex]\sigma[/itex]-compact space like the complex plane has to be regular then?
No. There are examples of Borel measures on [itex]\sigma[/itex]-compact (even on compact) spaces which fail to be regular.
 
  • #3
morphism said:
Can you explain how you're "regularizing" [itex]\mu[/itex]?
[itex]\mu[/itex] is associated with a bounded linear functional by [itex]\Phi(f) = \int_Xfd\mu[/itex]. Then by the Riesz representation theorem, if X is locally compact & Hausdorff, [itex]\Phi[/itex] is associated with a regular measure [itex]\mu^\prime[/itex] by [itex]\Phi(f) = \int_Xfd\mu^\prime[/itex]. So [itex]\int_Xfd(\mu-\mu^\prime)=0,[/itex] all [itex]f[/itex] in [itex]C_0(X)[/itex].
So the question is, how much does this say about the measure [itex]\mu-\mu^\prime[/itex]? Under what circumstances is it 0, so that all the complex measures on X are regular?
No. There are examples of Borel measures on [itex]\sigma[/itex]-compact (even on compact) spaces which fail to be regular.
If X is locally compact and Hausdorff can this still happen? example?
 
Last edited:
  • #4
The argument in Rudin's RACA for why the regular complex measure in the Riesz Representation theorem is unique, does apply just to regular measures. That's because the theorem that [itex]C_c(X)[/itex] (continuous fctns on X with compact support) is dense in [itex]L^p(\mu)[/itex] for [itex]1\le p [/itex]< [itex]\infty[/itex] just applies to regular measures.
From a theorem in Rudin's RACA, if [itex]X[/itex] is locally compact and Hausdorff, and every open set is [itex]\sigma[/itex]-compact, and [itex]\mu[/itex] is a complex Borel measure on [itex]X[/itex], [itex]|\mu|[/itex] is regular.
So I'm still wondering about regularizing complex measures, as described above, so that you get a new regular complex measure which gives the same integrals on [itex]C_0(X)[/itex]. How similar is the new measure? any examples of what happens with this process?
Laura
 
Last edited:
  • #5
lark said:
If X is locally compact and Hausdorff can this still happen? example?
The standard example is X=[0,w] where w is the first uncountable ordinal. This is an exercise in Rudin (last one in chapter 2 if you have the first edition).

As for your other question, I don't really know what happens in general. I'll think about it some more and let you know if I come up with anything.
 
  • #6
morphism said:
As for your other question, I don't really know what happens in general. I'll think about it some more and let you know if I come up with anything.

Yeah, I was wondering if regularizing a measure this way is something that's mathematically useful, that you would get a measure that gives the same integrals (a lot of the time at least) but is better behaved. In the usual sensible spaces, all the complex measures are regular anyway, so maybe the measures that aren't regular are mostly weird counterexamples that people don't care about regularizing.
Laura
 
  • #7
see attached .pdf
Laura
 

Attachments

  • junk.pdf
    29.5 KB · Views: 201

1. What are regularizing measures?

Regularizing measures refer to techniques used in statistics and machine learning to prevent overfitting and improve the generalization ability of a model by introducing additional constraints or penalties to the model parameters.

2. Why are regularizing measures important?

Regularizing measures are important because they help to prevent overfitting, which occurs when a model is overly complex and performs well on the training data but poorly on new data. Regularization helps to balance the trade-off between model complexity and generalization performance, resulting in more reliable and accurate models.

3. What are some common types of regularizing measures?

Some common types of regularizing measures include L1 and L2 regularization, dropout, early stopping, and data augmentation. L1 and L2 regularization add penalties to the model parameters to discourage large weights, while dropout randomly drops units from the neural network during training to prevent over-reliance on specific features. Early stopping stops the training process when the model's performance on a validation set stops improving, preventing overfitting. Data augmentation involves generating additional training data by applying transformations to the existing data.

4. How do regularizing measures work?

Regularizing measures work by adding constraints or penalties to the model parameters, forcing the model to learn more general patterns in the data instead of memorizing specific examples from the training set. This helps to prevent overfitting and improve the model's ability to generalize to new data.

5. When should regularizing measures be used?

Regularizing measures should be used when training a model that is prone to overfitting, such as deep neural networks with a large number of parameters. Regularization is especially important when working with small datasets, as these models are more likely to overfit due to the limited amount of data available for training.

Similar threads

Replies
3
Views
859
  • Science and Math Textbooks
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
667
Replies
3
Views
639
  • Science and Math Textbooks
Replies
7
Views
2K
Replies
1
Views
2K
  • Mechanics
Replies
31
Views
673
Replies
10
Views
924
Back
Top