A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral

By [L1], choose bounded Borel functions $\phi ,\psi$ on $[a,b]$ with [L1, L2] $\phi \le f\le \psi$ and both integrals equal to $I$. Then $0\le \psi -\phi$ and ${\int }_{[a,b]}(\psi -\phi )d{\lambda }_1=0.$ So [L2] gives $\psi =\phi$ almost everywhere. Since $\phi \le f\le \psi$, the same null set yields $f=\phi$ almost everywhere.

By [L4], the measure space $(\mathbb{R},\mathcal{L}(\mathbb{R}),{\lambda }_1)$ is [step 1.1, L3, L4] complete. The function $\phi$ is measurable because it is Borel, so [L3] applied to step 1.1 shows that $f$ is Lebesgue measurable.

The constant function $B{\chi }_{[a,b]}$ is a nonnegative simple measurable [step 2.1, L1, L6, L7] function, and [L6] together with [L7] gives ${\int }_{[a,b]}B{\chi }_{[a,b]}d{\lambda }_1=B(b-a)<+\mathrm{\infty }\mathrm{.}$ Since step 2.1 makes $\mathrm{\mid }f\mathrm{\mid }$ measurable and $\mathrm{\mid }f\mathrm{\mid }\le B{\chi }_{[a,b]}$, [L6] yields ${\int }_{[a,b]}\mathrm{\mid }f\mathrm{\mid }d{\lambda }_1<+\mathrm{\infty }\mathrm{.}$ Hence $f$ is Lebesgue integrable on $[a,b]$. The same estimate applies to $\phi$, because [L1] gives $\mathrm{\mid }\phi \mathrm{\mid }\le B$.

Steps 1.1 and 3.1 show that $f$ and $\phi$ are integrable and agree [step 1.1, step 3.1, L1, L5] almost everywhere. Taking the measurable set $A=[a,b]$ in [L5] gives ${\int }_{[a,b]}fd{\lambda }_1={\int }_{[a,b]}\phi d{\lambda }_1\mathrm{.}$ By [L1], the right-hand side is $I={\int }_a^{b}f(x)dx$. So the Lebesgue and Riemann integrals of $f$ agree. ∎