# Is there a sigma-algebra for the parameter space in relation to the likelihood function?

The likelihood function, $$L(\theta|x) = f(x|\theta)$$, is sometimes mistaken to be a pdf. I have always thought that showing an example where $$\int_{-\infty}^{\infty} L(\theta|x) d \theta = 1$$ does not hold would be enough to conclude that the likelihood function is not a pdf, and I have not given it much thought beyond that.

Recently, I read some posts 1) What is the difference between "likelihood" and "probability"?, 2) What is the reason that a likelihood function is not a pdf?, 3) How to rigorously define the likelihood?, and in the comments to an answer of 2) it is stated that

"The $$𝑑\theta$$ has not even a sense in general because there's not even a $$\sigma$$-field in the parameter space!"

Whereas I have no reason to doubt this, it is not obvious to me why it is true, probably because my familiarity with $$\sigma$$-algebras is limited to the one page in Casella & Berger's Statistical Inference, and the post Why do we need sigma-algebras to define probability spaces?.

Since the comment is more than 10 years old, I gathered it would make sense to post it as a question. There are more recent discussions in the comments of Intuition for why likelihood function sometimes *is* a PDF, not mentioning $$\sigma$$-algebra specifically, so perhaps there exists some other statement regarding the senselessness of $$d\theta$$ without involving $$\sigma$$-algebra, but I have a feeling they might be connected.

• The issue isn't whether there might exist some sigma-algebra on the parameter space $\Theta$ (trivially, there always exist many sigma-algebras). It's whether a sigma algebra has been defined by the modeler or is naturally determined by the model or its intended application. – whuber Mar 22 at 16:45
• @whuber Would it make sense to edit the question to "is there a naturally determined sigma-algebra...?" I guess the answer to that question is no ... – kajsam Mar 22 at 16:48
• $\int_{\Theta} L(\theta) \mathrm{d}\theta$ generally doesn't equal 1, but as long as it's not unbounded one should be able to define a normalizing constant. After all, the Gaussian integral $\int_{-\infty}^{\infty} \exp(-x^2) \mathrm{d}x$ only equals 1 when divided by $\sqrt{\pi}$, which then gives the normal density. – Durden Apr 14 at 2:36
• @Durden Yes, I'm well aware of that. But, as in the comment I've cited in my question, the $\sigma$-algebra question seem to precede the integration. From what I understand from whuber's comment, I believe that one is skipping a step (defining the $\sigma$-algebra) when directly going for the integration argument, even though it is very efficient, and also more accessible to most people not familiar with set theory. – kajsam Apr 16 at 11:02