Measurability with respect to completion

[haljordan45]

How can one show that a positive function with a Lebesgue integral is measurable with respect to the complete sigma algebra?

 

[AlexChandler]

Just as always, you must show that $$\{x : f(x)< \alpha \}$$ is a set in the sigma algebra for any $$\alpha \in \mathbb{R}.$$

 

[haljordan45]

Ok, but how does the Lebesgue integral aspect factor into the argument?

 

[AlexChandler]

Ok, but how does the Lebesgue integral aspect factor into the argument?

You need to show the function is measurable with respect to the lebesgue measure. Thus given $$\alpha \in \mathbb{R}$$ you must show that
$$\{x|f(x)< \alpha \}$$
is a lebesgue measurable set.

 

[AlexChandler]

Also, a couple of comments are in order regarding your initial post.

1) It doesn't make sense to ask whether a function is measurable with respect to a sigma algebra. What you should be asking is how to show that the function $$f$$ is measurable with respect to the measure space (which according to your post I can only assume is)
$$(\mathbb{R}, \mathcal{M} ,m) ,$$
that is, the real line together with the sigma algebra of all lebesgue measurable sets, and the lebesgue measure. Doing this is simply a matter of definition which I have given in the previous post.

2) It doesn't make sense to call a sigma algebra complete (unless this means something specific which I am unaware of). I can only assume the word complete here is referring to the fact that the space which I had previously mentioned is a complete measure space, that is for any subset $$E \in \mathcal{M}$$ and $$A \subseteq E$$ we have $$m(E)=0 \Rightarrow A \in \mathcal{M}.$$

Thus your post should have read:

How does one show a positive function is measurable with respect to the measure space $$(\mathbb{R}, \mathcal{M} ,m)$$