Difference between $L^{1}([0,1],\text{Borel},\lambda|_{\text{Borel}})$ and $L^{1}([0,1],\text{Lebesgue},\lambda)$

Question

In many applications, people simply equip $[0,1]$ with the Borel $\sigma$-algebra $\mathcal{B}$ and the restricted Lebesgue measure $\lambda|_{\mathcal{B}}$. Suppose that $\mathcal{L}$ denotes the Lebesgue $\sigma$-algebra on $[0,1]$ with Lebesgue measure $\lambda$. I don't understand how this affects integrals, i.e., what is the difference between $\int_{[0,1]}f \ d\lambda|_{\mathcal{B}}$ and $\int_{[0,1]}g \ d\lambda$ for $f\in L^{1}([0,1],\mathcal{B},\lambda|_{\mathcal{B}})$ and $g\in L^{1}([0,1],\mathcal{L},\lambda)$?

Answer 1

For $f\in \mathcal{L}^1([0,1], \mathcal{B}, \lambda_{|\mathcal{B}})$, let $[f]_{\mathcal{B}}$ denote the set of functions in $\mathcal{L}^1([0,1], \mathcal{B}, \lambda_{|\mathcal{B}})$ that agree $\lambda_{|\mathcal{B}}$-a.e. with $f$. Recall that $L^1([0,1], \mathcal{B}, \lambda_{|\mathcal{B}})=\{[f]_{\mathcal{B}}: f\in \mathcal{L}^1([0,1], \mathcal{B}, \lambda_{|\mathcal{B}})\}$ and is equipped with the norm $||f||_{\mathcal{B}}:=\int |f|$ $d\lambda_{|\mathcal{B}}$.

Similarly, if $\mathcal{M}$ denotes the Lebesgue measurable sets, for $f\in \mathcal{L}^1([0,1], \mathcal{M}, \lambda)$, let $[f]_{\mathcal{M}}$ denote the set of functions in $\mathcal{L}^1([0,1], \mathcal{M}, \lambda)$that agree $\lambda$-a.e. with $f$. Again, recall that $L^1([0,1], \mathcal{M}, \lambda)=\{[f]_{\mathcal{M}}: f\in \mathcal{L}^1([0,1], \mathcal{M}, \lambda)\}$ equipped with the norm $||f||_{\mathcal{M}}:=\int |f|$ $d\lambda$.

Now consider the obvious map $L^1([0,1], \mathcal{B}, \lambda_{|\mathcal{B}})\ni [f]_{\mathcal{B}}\mapsto [f]_{\mathcal{M}}\in L^1([0,1], \mathcal{M}, \lambda)$. It is well defined because if two functions in $\mathcal{L}^1([0,1], \mathcal{B}, \lambda_{\mathcal{|B}})$ agree $\lambda_{\mathcal{|B}}$-a.e., they certainly agree $\lambda$-a.e.. It is injective because if two functions in $\mathcal{L}^1([0,1], \mathcal{B}, \lambda_{|\mathcal{B}})$ agree $\lambda$-a.e., then, becuase their points of agreement/disagreement is a $\mathcal{B}-$measurable set, they agree $\lambda_{|\mathcal{B}}$-a.e.. The map is isometric because if $f\in \mathcal{L}^1([0,1], \mathcal{B}, \lambda_{|\mathcal{B}})$ there exists an increasing sequence of Borel measurable functions, $\{\sum\limits_{k=1}^{N_n}\alpha_{n,k}\mathcal{X}_{B_{n,k}}\}_n$ converging pointwise to $|f|$. Then $\int |f|\text{ }d\lambda_{|\mathcal{B}}=\lim\limits_n\sum\limits_{k=1}^{N_n}\alpha_{n,k}\lambda_{|\mathcal{B}}(B_{n,k})=\lim\limits_n \sum\limits_{k=1}^{N_n}\alpha_{n,k}\lambda(B_{n,k})=\int |f|\text{ }d\lambda$.

Here we have used the DCT.

So far we have only needed that $\mathcal{B} \subset \mathcal{M}$ so the above can clearly be generalized. But, what is special about $\mathcal{B}$ and $\mathcal{M}$ is that the map is surjective since any Lebesgue measurable function is $\lambda$-a.e. equal to a Borel measurable function. This shows that the two Banach spaces $L^1([0,1], \mathcal{B}, \lambda_{|\mathcal{B}})$ and $L^1([0,1], \mathcal{M}, \lambda)$ are canonically isometrically isomorphic, i.e. they are essentially the same.