Example of a Borel measurable function $f: [a,b] \to \Bbb{R}$ that is not integrable w.r.t. Borel sets but is w.r.t.Lebesgue measurable sets

Question

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.

Answer 1

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.

Answer 2

Consider the following

Lemma: Let $X$ be a set and $\mathcal{S} \subseteq \mathcal{T}$ two $\sigma$-algebras on $X$. Let $\mu_1$ be a measure on $\mathcal{T}$ and let $\mu_2$ be its restriction to $\mathcal{S}$. Then for every $\mathcal{S}$-measurable function $f: X \to [0, \infty]$ we have $$\int_X f d \mu_1 = \int_X f d \mu_2$$

Proof: For simple $\mathcal{S}$-measurable functions this is immediate. Then use that every positive $\mathcal{S}$-measurable is the monotone limit of simple functions. $\quad \square$

Now, apply the lemma with $\mathcal{S}$= Borel sets, $\mathcal{T}$ = Lebesgue measurable sets, $\mu_1 = \lambda$ and $\mu_2 = \lambda_b$ to conclude that the situation you are asking for cannot happen.