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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.

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    $\begingroup$ 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. $\endgroup$ – whuber Mar 22 at 16:45
  • $\begingroup$ @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 ... $\endgroup$ – kajsam Mar 22 at 16:48
  • $\begingroup$ $\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. $\endgroup$ – Durden Apr 14 at 2:36
  • $\begingroup$ @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. $\endgroup$ – kajsam Apr 16 at 11:02

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