The Lebesgue and Riemann Integrals Compared — Examples

By [L1], the function $f$ is $1$ on $\mathbb{Q}\cap [0,1]$ and $0$ on its [L1, L2, L3, L4] complement. The set $\mathbb{Q}\cap [0,1]$ is countable, hence null by [L2], and [L3] makes it Lebesgue measurable. Therefore $f=0$ almost everywhere on $[0,1]$, so [L4] gives ${\int }_{[0,1]}fd{\lambda }_1=0.$

By [L5], every point of $[0,1]$ is a discontinuity of $f$. Thus the [L5, L6, L7] discontinuity set of $f$ is the whole interval $[0,1]$, whose Lebesgue measure is $1$ by [L7], not $0$. So [L6] shows that $f$ is not Riemann integrable. ∎

By [L1], the compact set $K$ has positive measure, so [L2] provides a [L1, L2, L3, L4, construct] nonmeasurable subset $N\subseteq K$. Put $E:={\psi }^{-1}[N]\subseteq C$. Since $C$ is measurable and null by [L3], step 1.1 and [L4] show that $E$ is Lebesgue measurable and null.

Let $h:={\mathbf{1}}_E$ on $[0,1]$. If $x\mathrm{\notin }C$, then [step 1.1, L3, L5, L6] an open neighbourhood of $x$ disjoint from $C$, hence disjoint from $E$, so $h$ is identically $0$ there. Thus every discontinuity of $h$ lies in $C$. Since $C$ has Lebesgue measure $0$ by [L3], [L6] makes $h$ Riemann integrable on $[0,1]$.

Suppose instead that $h$ were Borel measurable on $[0,1]$. [step 1.1, step 2.1, L1, L7, L8, assume-contra, discharge-contradiction] Then $E=h^{-1}((1\mathrm{/}2,\mathrm{\infty }))$ would be Borel in the subspace $[0,1]$. Because $E\subseteq C$, [L7] makes $E$ Borel in the subspace $C$. The restriction $\psi {\mathrm{\mid }}_C:C\to K$ is a homeomorphism by [L1], so its inverse $(\psi {\mathrm{\mid }}_C{)}^{-1}:K\to C$ is continuous; applying [L8] to that inverse shows that $N=\psi [E]$ is Borel in the subspace $K$. Since $K$ is compact and therefore closed in $\mathbb{R}$, [L7] makes $N$ Borel in $\mathbb{R}$, contradicting the choice of $N$. Therefore $h$ is Riemann integrable but not Borel measurable, and the statement is false. ∎

By [L1], fix an enumeration $(q_k{)}_{k\ge 0}$ of $\mathbb{Q}\cap (0,1)$. For each $k\ge 0$, let $I_k:=(q_k-2^{-k-4},q_k+2^{-k-4})\cap (0,1),$ put $U_n:={\bigcup }_{k<n}{I}_k$ for $n\ge 1$, and put $U:={\bigcup }_{k\ge 0}{I}_k$. Each $U_n$ and $U$ is open, and every rational point of $(0,1)$ lies in $U$, so $U$ is dense in $(0,1)$ and hence in $[0,1]$. Each interval $I_k$ has length at most $2^{-k-3}$, so [L2], [L3], [L8], and [L9] give ${\lambda }_1(U)\le {\sum }_{k=0}^{\mathrm{\infty }}{\lambda }_1(I_k)\le {\sum }_{k=0}^{\mathrm{\infty }}{2}^{-k-3}=2^{-3}{\sum }_{k=0}^{\mathrm{\infty }}{2}^{-k}=1\mathrm{/}4<1.$

Each $U_n$ is a finite union of open intervals, so ${\mathbf{1}}_{U_n}$ is continuous away from the finitely many endpoints of those intervals. Thus [L5] makes every ${\mathbf{1}}_{U_n}$ Riemann integrable on $[0,1]$. Also $U_n\uparrow U$, so ${\mathbf{1}}_{U_n}\uparrow {\mathbf{1}}_U$ pointwise and $\mathrm{\mid }{\mathbf{1}}_U-{\mathbf{1}}_{U_n}\mathrm{\mid }={\mathbf{1}}_{U\setminus U_n}\mathrm{.}$ By [L4] and [L7], ${\int }_{[0,1]}\mathrm{\mid }{\mathbf{1}}_U-{\mathbf{1}}_{U_n}\mathrm{\mid }d{\lambda }_1={\lambda }_1(U\setminus U_n)={\lambda }_1(U)-{\lambda }_1(U_n)\to 0.$

Let $C:=[0,1]\setminus U$. Step 1.1 and [L8] give ${\lambda }_1(C)={\lambda }_1([0,1])-{\lambda }_1(U)=1-{\lambda }_1(U)>0.$ Because $U$ is open, every point of $U$ is a continuity point of ${\mathbf{1}}_U$. Because $U$ is dense, every point of $C$ is a boundary point of $U$, hence every neighbourhood of such a point meets both $U$ and $C$; so ${\mathbf{1}}_U$ is discontinuous at every point of $C$. Therefore the discontinuity set of ${\mathbf{1}}_U$ contains the positive-measure set $C$, and [L6] shows that ${\mathbf{1}}_U$ is not Riemann integrable on $[0,1]$.

The function $f$ is continuous on $[0,1]$, so its proper integral there exists. By [L1], the improper integral ${\int }_1^{\mathrm{\infty }}\frac{\sin x}{x}dx$ converges. For $R>1$, [L3] gives ${\int }_0^{R}f(x)dx={\int }_0^{1}f(x)dx+{\int }_1^R\frac{\sin x}{x}dx\mathrm{.}$ Passing to the limit as $R\to \mathrm{\infty }$ and using [L4], the improper integral ${\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}dx$ converges. In the terminology of [L6], the sine integral is convergent.

Apply [L2] with the nonincreasing function $g(x)=1\mathrm{/}x$, the bounded-gap sequence $x_j=j\pi$, and the locally integrable oscillatory factor $u(x)=\sin x$. Since ${\int }_{j\pi }^{(j+1)\pi }\mathrm{\mid }\sin x\mathrm{\mid }dx=2(j\ge 1),$ [L2] yields divergence of ${\int }_1^{\mathrm{\infty }}\frac{\mathrm{\mid }\sin x\mathrm{\mid }}{x}dx\mathrm{.}$ Because $\mathrm{\mid }f\mathrm{\mid }$ is continuous on $[0,1]$, its proper integral there exists, so [L3] and [L4] show that ${\int }_0^{\mathrm{\infty }}\mathrm{\mid }f(x)\mathrm{\mid }dx$ also diverges. Thus the improper integral of $\sin x\mathrm{/}x$ is not absolutely convergent.

By [L6], steps 1.1 and 1.2 show that the sine integral is only conditionally convergent. If $f$ were Lebesgue integrable on $[0,\mathrm{\infty })$, then [L5] would force ${\int }_{[0,\mathrm{\infty })}\mathrm{\mid }f\mathrm{\mid }d{\lambda }_1<+\mathrm{\infty },$ so $\mathrm{\mid }f\mathrm{\mid }$ would be Lebesgue integrable on the half-line. For each natural number $n\ge 1$, put $h_n:=\mathrm{\mid }f\mathrm{\mid }{\chi }_{[0,n]}\mathrm{.}$ Then $0\le h_n\le h_{n+1}$ and $h_n(x)\uparrow \mathrm{\mid }f(x)\mathrm{\mid }$ for every $x\ge 0$. Since $\mathrm{\mid }f\mathrm{\mid }$ is continuous on $[0,n]$, [L7] makes it Riemann integrable there, and [L8] gives ${\int }_{[0,\mathrm{\infty })}{h}_{n}d{\lambda }_1={\int }_{[0,n]}\mathrm{\mid }f\mathrm{\mid }d{\lambda }_1={\int }_0^n\mathrm{\mid }f(x)\mathrm{\mid }dx\mathrm{.}$ Because $h_n\uparrow \mathrm{\mid }f\mathrm{\mid }$, [L9] yields ${\lim }_{n\to \mathrm{\infty }}{\int }_0^n\mathrm{\mid }f(x)\mathrm{\mid }dx={\lim }_{n\to \mathrm{\infty }}{\int }_{[0,\mathrm{\infty })}{h}_{n}d{\lambda }_1={\int }_{[0,\mathrm{\infty })}\mathrm{\mid }f\mathrm{\mid }d{\lambda }_1<+\mathrm{\infty }\mathrm{.}$ By [L4], that finite limit is exactly the improper integral ${\int }_0^{\mathrm{\infty }}\mathrm{\mid }f(x)\mathrm{\mid }dx,$ so the absolute improper integral converges, contradicting step 1.2. Therefore $f$ is not Lebesgue integrable on the half-line, and the statement is false.

The function ${\chi }_S$ is upper semicontinuous on $[0,1]$. If $x\in S$, then ${\chi }_S(y)\le 1<1+\epsilon$ for every $y$ and every $\epsilon >0$. If $x\mathrm{\notin }S$, then $S$ is closed by [L2], so there is an open neighbourhood of $x$ disjoint from $S$; on that neighbourhood ${\chi }_S$ is identically $0$, which is certainly below $0+\epsilon$. Thus [L1] is satisfied at every point.

Let $g:[0,1]\to \mathbb{R}$ satisfy $g={\chi }_S$ almost everywhere, and let $N$ be a measurable null set containing the disagreement set. Because $S$ is compact, [L3] makes it measurable, and [L2] says it does not have measure zero. Therefore [L4] gives ${\lambda }_1(S\setminus N)={\lambda }_1(S)-{\lambda }_1(S\cap N)={\lambda }_1(S)>0.$ Fix $x\in S\setminus N$. Then $g(x)=1$. Since $S$ is nowhere dense by [L2], every neighbourhood of $x$ meets $[0,1]\setminus S$; the intersection of that neighbourhood with $[0,1]\setminus S$ is a nonempty open set, so it contains a nondegenerate interval and hence has positive measure by [L6]. Because $N$ is null, that positive-measure open set cannot be contained in $N$, so the neighbourhood contains $y\mathrm{\notin }S\cup N$. Then $g(y)={\chi }_S(y)=0$. Thus $g$ is discontinuous at every $x\in S\setminus N$.

Step 2.1 shows that the discontinuity set of $g$ contains the positive-measure set $S\setminus N$. Hence [L5] implies that $g$ is not Riemann integrable. So ${\chi }_S$ is upper semicontinuous but is almost everywhere equal to no Riemann integrable function, and the statement is false.

By [L1], fix an enumeration $(q_j{)}_{j\ge 0}$ of $\mathbb{Q}$. For each natural number $k\ge 1$, put $U_k:={\bigcup }_{j\ge 0}(q_j-2^{-j-k-2},q_j+2^{-j-k-2})\mathrm{.}$ Every $U_k$ is open and dense, because it contains every rational. Each interval in the union has length $2^{-j-k-1}$, so [L2], [L3], and [L10] give ${\lambda }_1(U_k)\le {\sum }_{j=0}^{\mathrm{\infty }}{2}^{-j-k-1}=2^{-k-1}{\sum }_{j=0}^{\mathrm{\infty }}{2}^{-j}=2^{-k}\mathrm{.}$ Let $G:={\bigcap }_{k\ge 1}{U}_k\mathrm{.}$ Then [L4] makes $G$ a $G_{\delta }$ set, and [L6] makes it dense. Since $G\subseteq U_k$ for every $k$, [L9] gives ${\lambda }_1(G)\le 2^{-k}$ for all $k$, so ${\lambda }_1(G)=0.$

Suppose, for contradiction, that $G$ is the discontinuity set of some function $f:\mathbb{R}\to \mathbb{R}$. Then [L5] makes $G$ an $F_{\sigma }$ set, so by [L4] write $G={\bigcup }_{n\ge 0}{F}_n$ with each $F_n$ closed. Because $F_n\subseteq G$ and ${\lambda }_1(G)=0$, monotonicity [L9] gives ${\lambda }_1(F_n)=0$ for every $n$. A closed null set cannot contain a nondegenerate interval, by [L8], so each $F_n$ has empty interior and is nowhere dense by [L7]. Therefore $G$ is meager.

Step 1.1 writes $G$ as a dense $G_{\delta }$ set, so $\mathbb{R}\setminus G$ is meager by [L7]. If $G$ were also meager, then $\mathbb{R}=G\cup (\mathbb{R}\setminus G)$ would be a union of two meager sets and therefore meager, contradicting [L6]. So $G$ is a Lebesgue null set that is not the discontinuity set of any real-valued function, and the statement is false.