Let $\lambda$ be Lebesgue measure on the Lebesgue measurable subsets of $\Bbb{R}$ and $\lambda_b$ its restriction to the Borel sets.
I'm looking for a Borel-measurable function $f: [a,b]\to [0, \infty]$ such that
$$\int_a^b f d \lambda < \infty$$
but $$\int_a^b f d \lambda_b = \infty$$
Does such a function exist? I guess not, but couldn't prove.
Best Answer
Aren't all Borel sets also Lebesgue measurable? The definitions of those integrals are: $$\int_{[a,b]}fd\lambda = \sup_{\phi}\bigg\{\int_{[a,b]}\phi\text{ } d\lambda :\phi\text{ is simple }, \phi \leq f \bigg\} $$ (Assuming f is non-negative, otherwise you break the integral apart into $f^+$ and $f^-$ but I'll assume non-negative for simplicity)
And similarly for $d\lambda_b$, the simple functions for $d\lambda_b$ are required to have their domain restricted to Borel sets, while the above integral allows for simple functions to vary over all lebesgue measurable sets. This implies that the set you're taking the supremum over for $\lambda_b$ is a subset of the one for $\lambda$, this implies that: $$\int_{[a,b]}fd\lambda_b = \sup_{\phi}\bigg\{\int_{[a,b]}\phi\text{ } d\lambda_b :\phi\text{ is simple }, \phi \leq f \bigg\} \leq \sup_{\phi}\bigg\{\int_{[a,b]}\phi\text{ } d\lambda :\phi\text{ is simple }, \phi \leq f \bigg\} = \int_{[a,b]}fd\lambda$$ So if the normal lebesgue integral is finite the other one has to be.