A bounded Riemann integrable function admits Borel Darboux envelopes with the same Lebesgue integral

By [L1], choose recursively a refining sequence of partitions $P_1⪯P_2⪯\cdots$ of $[a,b]$ such that $U(f,P_n)-L(f,P_n)<1\mathrm{/}n$ for every $n\ge 1$: choose $P_1$ for $\epsilon =1$, and once $P_n$ is chosen, let $R_{n+1}$ satisfy $U(f,R_{n+1})-L(f,R_{n+1})<1\mathrm{/}(n+1)$ and put $P_{n+1}:=P_n\vee R_{n+1}$; then [L2] preserves the inequality under refinement. For each $n$, write $P_n=\{a=t_0^{(n)}<\cdots <t_{m_n}^{(n)}=b\}$ and let $m_{n,i}$ and $M_{n,i}$ be the infimum and supremum of $f$ on $[t_i^{(n)},t_{i+1}^{(n)}]$. Define ${\mathrm{\ell }}_n:=m_{n,m_n-1}{\chi }_{\{b\}}+{\sum }_{i<m_n}{m}_{n,i}{\chi }_{[t_i^{(n)},t_{i+1}^{(n)})},$ $u_n:=M_{n,m_n-1}{\chi }_{\{b\}}+{\sum }_{i<m_n}{M}_{n,i}{\chi }_{[t_i^{(n)},t_{i+1}^{(n)})}\mathrm{.}$ Each partition piece is a Borel subset of $[a,b]$ by [L3], so ${\mathrm{\ell }}_n$ and $u_n$ are bounded Borel functions on $[a,b]$. Also $-B\le {\mathrm{\ell }}_n\le {\mathrm{\ell }}_{n+1}\le f\le u_{n+1}\le u_n\le B$ pointwise, while [L11] gives $L(f,P_n)\le I\le U(f,P_n),U(f,P_n)-L(f,P_n)<1\mathrm{/}n\mathrm{.}$

Put $\phi :={\sup }_n{\mathrm{\ell }}_n$ and $\psi :={\inf }_n{u}_n$. Since ${\mathrm{\ell }}_n\uparrow \phi$, the last clause of [L4] makes $\phi$ Borel measurable; since $u_n$ are measurable, the infimum clause of [L4] makes $\psi$ Borel measurable. Step 1.1 gives $-B\le \phi \le f\le \psi \le B$. Now $B+{\mathrm{\ell }}_n$ is a nonnegative simple function, so [L6] and [L8] give ${\int }_{[a,b]}(B+{\mathrm{\ell }}_n)d{\lambda }_1=B(b-a)+L(f,P_n)\mathrm{.}$ Because $B+{\mathrm{\ell }}_n\uparrow B+\phi$, [L7] yields ${\int }_{[a,b]}(B+\phi )d{\lambda }_1={\lim }_{n\to \mathrm{\infty }}(B(b-a)+L(f,P_n))=B(b-a)+I,$ the limit being the squeeze from step 1.1. Since $\mathrm{\mid }(B+\phi )-B{\chi }_{[a,b]}\mathrm{\mid }=\mathrm{\mid }\phi \mathrm{\mid }\le B{\chi }_{[a,b]}$, the constant function $B{\chi }_{[a,b]}$ is integrable by [L6] and [L8], so step 1.1 and [L5], [L9] show that $\phi \in L^1({\lambda }_1)$ and ${\int }_{[a,b]}\phi d{\lambda }_1=I\mathrm{.}$

For each $n$ the function $u_n-{\mathrm{\ell }}_n$ is nonnegative simple, and step 1.1 with [L6] and [L8] gives ${\int }_{[a,b]}(u_n-{\mathrm{\ell }}_n)d{\lambda }_1=U(f,P_n)-L(f,P_n)<1\mathrm{/}n\mathrm{.}$ Because $\phi \le f\le \psi$ and ${\mathrm{\ell }}_n\le \phi \le \psi \le u_n$, one has $0\le \psi -\phi \le u_n-{\mathrm{\ell }}_n$ for every $n$. So [L5] yields $0\le {\int }_{[a,b]}(\psi -\phi )d{\lambda }_1<1\mathrm{/}n(n\ge 1)\mathrm{.}$ If that integral were positive, [L10] would give $n$ with $1\mathrm{/}n<{\int }_{[a,b]}(\psi -\phi )d{\lambda }_1$, contradicting the displayed inequality. Therefore ${\int }_{[a,b]}(\psi -\phi )d{\lambda }_1=0.$ The same bound $\mathrm{\mid }\psi -\phi \mathrm{\mid }\le 2B{\chi }_{[a,b]}$ shows $\psi -\phi \in L^1({\lambda }_1)$.

Since $\psi =\phi +(\psi -\phi )$ and both summands are integrable, [L9] and step 3.1 give ${\int }_{[a,b]}\psi d{\lambda }_1={\int }_{[a,b]}\phi d{\lambda }_1+{\int }_{[a,b]}(\psi -\phi )d{\lambda }_1=I+0=I\mathrm{.}$ Together with steps 2.1 and 3.1, this proves the existence of bounded Borel envelopes $\phi \le f\le \psi$ with the same Lebesgue integral $I$.