The Lebesgue and Riemann Integrals Compared

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 $I$.

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. ∎

By [L1], choose bounded Borel functions $\phi ,\psi$ with [L1, L2] $\phi \le f\le \psi$ and ${\int }_{[a,b]}(\psi -\phi )d{\lambda }_1=0.$ Since $\psi -\phi \ge 0$, [L2] gives $\psi =\phi$ almost everywhere on $[a,b]$.

On the same full-measure set one has [step 1.1, L1] $\phi \le f\le \psi =\phi$, so $f=\phi$ almost everywhere. Taking $g:=\phi$ proves the claim, and $g$ is Borel by step 1.1. ∎

By [L1], each $f_n$ and $f$ is Lebesgue measurable and integrable on $[a,b]$, and ${\int }_{[a,b]}{f}_{n}d{\lambda }_1={\int }_a^b{f}_n(x)dx,{\int }_{[a,b]}fd{\lambda }_1={\int }_a^{b}f(x)dx\mathrm{.}$ Also [L3] makes $([a,b],\mathcal{L}([a,b]),{\lambda }_1)$ a finite measure space.

The convergence hypothesis is pointwise, hence almost everywhere, and the uniform bound $\mathrm{\mid }{f}_n\mathrm{\mid }\le M$ holds everywhere. So [L2] applies on $[a,b]$ and gives ${\int }_{[a,b]}{f}_{n}d{\lambda }_1⟶{\int }_{[a,b]}fd{\lambda }_1\mathrm{.}$ Translating the two sides with step 1.1 yields the stated convergence of the Riemann integrals.

For each natural number $n\ge 1$, define $f_n:=f{\chi }_{[a,a+n]}\mathrm{.}$ Then $0\le f_n\le f_{n+1}$ and $f_n(x)\uparrow f(x)$ for every $x\ge a$. By [L1] applied on $[a,a+n]$, ${\int }_{[a,\mathrm{\infty })}{f}_{n}d{\lambda }_1={\int }_{[a,a+n]}fd{\lambda }_1={\int }_a^{a+n}f(x)dx\mathrm{.}$

Since $f_n\uparrow f$, [L2] gives ${\int }_{[a,\mathrm{\infty })}fd{\lambda }_1={\lim }_{n\to \mathrm{\infty }}{\int }_{[a,\mathrm{\infty })}{f}_{n}d{\lambda }_1={\lim }_{n\to \mathrm{\infty }}{\int }_a^{a+n}f(x)dx\mathrm{.}$ Because $a+n\to +\mathrm{\infty }$, [L3] identifies the last limit with the given improper integral $L$. Hence ${\int }_{[a,\mathrm{\infty })}fd{\lambda }_1=L,$ and in particular the Lebesgue integral is finite, so $f$ is integrable on the half-line.

Extend $g$ to a continuous function $\tilde{g}:\mathbb{R}\to \mathbb{R}$ by setting $\tilde{g}(x)=g(a)$ for $x<a$, $\tilde{g}(x)=g(x)$ for $x\in [a,b]$, and $\tilde{g}(x)=g(b)$ for $x>b$. Then [L5] makes $\tilde{g}$ Borel measurable. Put $h:=(B+\tilde{g}){\chi }_{(a,b]}\mathrm{.}$ This is a nonnegative measurable function. Since $0\le h\le 2B{\chi }_{(a,b]}$, [L3], [L4], [L6], and [L7] show that $h$ is ${\mu }_F$-integrable. Let $J:={\int }_a^b(B+g)dF\mathrm{.}$ By [L1], the integral $J$ exists. The constant integrand $B$ has the same Riemann-Stieltjes sum $B(F(b)-F(a))$ for every tagged partition, so [L8] gives ${\int }_a^{b}BdF=B(F(b)-F(a))\mathrm{.}$ Therefore [L9] yields $J=I+B(F(b)-F(a))\mathrm{.}$

Let $\epsilon >0$. By [L2], choose a partition $P_0=\{a=t_0<\cdots <t_m=b\}$ with $U_F(g,P_0)-L_F(g,P_0)<\epsilon$. By [L8], choose $\delta >0$ such that every tagged partition of mesh below $\delta$ has Riemann-Stieltjes sum for $B+g$ within $\epsilon$ of $J$. Let $P=\{a=s_0<\cdots <s_n=b\}$ be a refinement of $P_0$ with mesh below $\delta$. For each $i$ put $m_i:={\inf }_{[s_{i-1},s_i]}g,M_i:={\sup }_{[s_{i-1},s_i]}g,$ and define nonnegative simple functions on $\mathbb{R}$ by ${\mathrm{\ell }}_P:={\sum }_{i=1}^n(B+m_i){\chi }_{(s_{i-1},s_i]},u_P:={\sum }_{i=1}^n(B+M_i){\chi }_{(s_{i-1},s_i]}\mathrm{.}$ Then ${\mathrm{\ell }}_P\le h\le u_P$. Because each refined infimum is at least the corresponding coarse infimum and each refined supremum is at most the corresponding coarse supremum, the Stieltjes lower sum increases and the upper sum decreases under this refinement, so $U_F(g,P)-L_F(g,P)\le U_F(g,P_0)-L_F(g,P_0)<\epsilon \mathrm{.}$ Also [L3] and [L4] give $\int {\mathrm{\ell }}_{P}d{\mu }_F=B(F(b)-F(a))+L_F(g,P),\int u_{P}d{\mu }_F=B(F(b)-F(a))+U_F(g,P)\mathrm{.}$

Fix any tagging $\xi$ of $P$. Then $\int {\mathrm{\ell }}_{P}d{\mu }_F\le \int hd{\mu }_F\le \int u_{P}d{\mu }_F,$ and because $m_i\le g({\xi }_i)\le M_i$ on every subinterval, $\int {\mathrm{\ell }}_{P}d{\mu }_F\le S(B+g,F;P,\xi )\le \int u_{P}d{\mu }_F\mathrm{.}$ Hence both $\int hd{\mu }_F$ and $S(B+g,F;P,\xi )$ lie in an interval of length $U_F(g,P)-L_F(g,P)<\epsilon$, so $\mid \int hd{\mu }_F-S(B+g,F;P,\xi )\mid <\epsilon \mathrm{.}$

Because $\mathrm{\parallel }P\mathrm{\parallel }<\delta$, [L8] gives $\mid S(B+g,F;P,\xi )-J\mid <\epsilon \mathrm{.}$ Combining this with step 2.1, $\mid \int hd{\mu }_F-J\mid <2\epsilon \mathrm{.}$ Since $\epsilon >0$ was arbitrary, $\int hd{\mu }_F=J=I+B(F(b)-F(a))\mathrm{.}$

Step 1.1 gives $h=\tilde{g}{\chi }_{(a,b]}+B{\chi }_{(a,b]}$, and both summands are integrable by step 1.1. Therefore [L3] and [L6] yield ${\int }_{(a,b]}gd{\mu }_F=\int hd{\mu }_F-B{\mu }_F((a,b])=(I+B(F(b)-F(a)))-B(F(b)-F(a))=I\mathrm{.}$ This is exactly ${\int }_{(a,b]}gd{\mu }_F={\int }_a^{b}gdF,$ so the two integrals agree.