One-dimensional $W^{1,p}$ functions have unique absolutely continuous representatives

By [F21] the given Axiom of Choice [F22] yields Countable and Dependent Choice, so the choice hypotheses of the ACL characterisation [F2], of the absolutely-continuous fundamental theorem [F5], of the first fundamental theorem [F8], of the measure [F11] and of the characterisation of Lebesgue measure on boxes [F10] are available. In particular $Du$ is an $L^p$ class by [F1], and it makes sense to integrate $Du$ over intervals.

Uniqueness claim on any nonempty open interval $J\subseteq \mathbb{R}$. Suppose $v$ and $w$ are representatives of one and the same almost-everywhere class on $J$ that are both absolutely continuous on every compact subinterval of $J$ and both satisfy the displayed formula with the same weak derivative class: $v(y)-v(x)={\int }_x^{y}Du$ and $w(y)-w(x)={\int }_x^{y}Du$ for all $x\le y$ in $J$. Then $h:=v-w$ satisfies $h(y)-h(x)=0$ for all $x\le y$, so $h$ is constant on $J$; and since $v=w$ almost everywhere on $J$, that constant is zero: otherwise $\{h\mathrm{\ne }0\}=J$ would be a null set, whereas $J$ contains a compact interval $[x,y]$ with $x<y$, whose measure $y-x$ is positive by [F10] and which therefore has measure at most $\lambda (J)$ by [F12] with $\lambda$ the measure of [F11]. Hence $v=w$ on $J$.

The finite-exponent case $1\le p<\mathrm{\infty }$. By [F2] applied to the class $u$ on the open interval $I$ there is a measurable ACL representative $u^{\ast }$ whose classical derivative $(u^{\ast }{)}^{\mathrm{\prime }}$ exists almost everywhere, is measurable, lies in $L^p(I)$, and represents $Du$ almost everywhere. By [F3] the representative $u^{\ast }$ is absolutely continuous on every compact subinterval of $I$. Fix $x\le y$ in $I$; the interval $[x,y]$ is compact in $I$. For real scalars [F5] applied to $u^{\ast }$ on $[x,y]$ gives $u^{\ast }(y)-u^{\ast }(x)={\int }_x^y(u^{\ast }{)}^{\mathrm{\prime }}$, and $(u^{\ast }{)}^{\mathrm{\prime }}=Du$ almost everywhere on $[x,y]$, so [F6] replaces the integrand and yields $u^{\ast }(y)-u^{\ast }(x)={\int }_x^{y}Du$. For complex scalars the same argument applied to the real and imaginary parts, which are absolutely continuous by [F3] and whose derivatives represent $\mathrm{Re}Du$ and $\mathrm{Im}Du$ almost everywhere, and the componentwise integration of [F16] give the same identity.

The case $p=\mathrm{\infty }$. Put $J_n:=I\cap (-n,n)$ for $n\ge 1$; each $J_n$ is an open interval with union $I$, and for nonempty $J_n$ the box measure formula [F10] gives $\lambda (J_n)<\mathrm{\infty }$. By [F1], $u$ and $Du$ are essentially bounded classes on $I$, hence on $J_n$; applying [F9] to the real measurable functions $\mathrm{\mid }u\mathrm{\mid }$ and $\mathrm{\mid }Du\mathrm{\mid }$ on the finite-measure space $J_n$ shows that $\mathrm{\mid }u\mathrm{\mid }$ and $\mathrm{\mid }Du\mathrm{\mid }$ lie in ${\mathcal{L}}^1(J_n)$, and the modulus convention of [F16] converts this into $u{\mathrm{\mid }}_{J_n}\in L^1(J_n;\mathbb{K})$ and $Du{\mathrm{\mid }}_{J_n}\in L^1(J_n;\mathbb{K})$. By [F4] the class $Du{\mathrm{\mid }}_{J_n}$ is the weak derivative of $u{\mathrm{\mid }}_{J_n}$ on $J_n$, so [F1] gives $u{\mathrm{\mid }}_{J_n}\in W^{1,1}(J_n;\mathbb{K})$.

The case $p=\mathrm{\infty }$, representatives. Fix $n$ with $J_n\mathrm{\ne }\mathrm{\varnothing }$ and apply the finite-exponent result of step 2.1 with exponent $1$ on the interval $J_n$ to the class $u{\mathrm{\mid }}_{J_n}\in W^{1,1}(J_n;\mathbb{K})$: there is a measurable representative $u_n$ of that class, absolutely continuous on every compact subinterval of $J_n$, with $u_n(y)-u_n(x)={\int }_x^{y}Du$ for all $x\le y$ in $J_n$. The class $u{\mathrm{\mid }}_{J_n}$ determines the set of admissible representatives $u_n$, and selecting one for each $n\ge 1$ is a countable selection, licensed by Countable Choice from step 1.1. If $J_m\cap J_n\mathrm{\ne }\mathrm{\varnothing }$, then on that nonempty open interval the two representatives represent the same class and both satisfy the formula there, so step 1.2 gives $u_m=u_n$ on $J_m\cap J_n$. Define $u^{\ast }(x):=u_n(x)$ for the least $n$ with $x\in J_n$; the agreement on overlaps makes this well defined, and $u^{\ast }{\mathrm{\mid }}_{J_n}=u_n$ for every $n$. Every compact interval $K\subseteq I$ is contained in some $J_n$: if $K\subseteq [x,y]$ with $x\le y$ in $I$, choose $n$ with $n>\max (\mathrm{\mid }x\mathrm{\mid },\mathrm{\mid }y\mathrm{\mid })$. Hence $u^{\ast }$ is absolutely continuous on every compact subinterval of $I$ and satisfies $u^{\ast }(y)-u^{\ast }(x)={\int }_x^{y}Du$ for all $x\le y$ in $I$.

Combining steps 2.1 and 3.1, in both the finite and the infinite case the class $u$ has a representative $u^{\ast }$ that is absolutely continuous on every compact subinterval of $I$ and satisfies $u^{\ast }(y)-u^{\ast }(x)={\int }_x^{y}Du$ for all $x\le y$ in $I$; equivalently the displayed identity $u^{\ast }(x)-u^{\ast }(y)={\int }_y^{x}Du$ holds for all $x,y\in I$ with the stated sign convention. Any other representative $v$ with the same two properties satisfies the hypotheses of step 1.2 on the interval $J:=I$, so $v=u^{\ast }$. Thus $u^{\ast }$ is the unique representative of the class with these properties.

The representative $u^{\ast }$ is continuous on $I$. Fix $x\in I$ and $\epsilon >0$, and choose $c<x<d$ with $[c,d]\subseteq I$. On the compact interval $[c,d]$ the class $Du$ lies in ${\mathcal{L}}^1$: for finite $p$ this is Holder [F13] applied to $\mathrm{\mid }Du\mathrm{\mid }$ and the constant function $1$, and for $p=\mathrm{\infty }$ it is [F9] on the finite-measure interval. So [F15] applied to $Du$ on $[c,d]$ gives $\delta >0$ such that ${\int }_E\mathrm{\mid }Du\mathrm{\mid }<\epsilon$ for every measurable $E\subseteq [c,d]$ with $\lambda (E)<\delta$. If $y\in [c,d]$ satisfies $\mathrm{\mid }y-x\mathrm{\mid }<\delta$, then after swapping $x$ and $y$ if necessary the interval $[x,y]\subseteq [c,d]$ has measure $\mathrm{\mid }y-x\mathrm{\mid }<\delta$ by [F10], and the formula of step 4.1 with [F14] gives $\mathrm{\mid }{u}^{\ast }(y)-u^{\ast }(x)\mathrm{\mid }=\mathrm{\mid }{\int }_x^{y}Du\mathrm{\mid }\le {\int }_x^y\mathrm{\mid }Du\mathrm{\mid }={\int }_{[x,y]}\mathrm{\mid }Du\mathrm{\mid }<\epsilon$. This is exactly the continuity condition at $x$: for real scalars [F18], and for complex scalars [F19] applied to ${\mathbb{R}}^2$ under the identification of [F20]. Hence $u^{\ast }$ is continuous on $I$.

The finite-endpoint extension. Assume $I=(a,b)$ with $-\mathrm{\infty }<a<b<\mathrm{\infty }$. Then $\lambda ((a,b))=b-a<\mathrm{\infty }$ by [F10], and $Du\in {\mathcal{L}}^1(a,b)$ in the same way as in step 5.1; fix $c_0\in (a,b)$ and define $H(x):={\int }_a^{x}Du$ for $x\in [a,b]$ and the extension $E:=H+u^{\ast }(c_0)-H(c_0)$ on $[a,b]$. By [F7] the function $H$ is absolutely continuous on $[a,b]$ (for complex scalars apply [F7] to $\mathrm{Re}Du$ and $\mathrm{Im}Du$ and use [F16]), and adding the constant $u^{\ast }(c_0)-H(c_0)$ preserves absolute continuity, so $E$ is absolutely continuous. The countable additivity of the integral over disjoint measurable sets [F17], together with the almost-everywhere agreement of the indicator of the union with the sum of the two indicators [F6], gives $H(x)-H(c_0)={\int }_{c_0}^{x}Du$ for every $x\in [a,b]$; hence $E(x)=u^{\ast }(c_0)+{\int }_{c_0}^{x}Du=u^{\ast }(x)$ for $x\in (a,b)$ by the formula of step 4.1, so $E$ extends $u^{\ast }$. Also $E(x)-E(a)={\int }_a^{x}Du$ for every $x\in [a,b]$. Finally, if $F$ is any absolutely continuous function on $[a,b]$ extending $u^{\ast }$, then [F5] gives $F(x)-F(a)={\int }_a^x{F}^{\mathrm{\prime }}$ for all $x$, and $F^{\mathrm{\prime }}=E^{\mathrm{\prime }}=Du$ almost everywhere on $(a,b)$ because $F$ and $E$ agree identically there; [F6] then gives $F(x)-F(a)={\int }_a^{x}Du=E(x)-E(a)$, so $F-E$ is constant on $[a,b]$, and that constant is $0$ because $F=E$ on $(a,b)$. Thus the extension is unique.

Steps 2.1 to 3.1 produce the representative $u^{\ast }$, step 4.1 states that it is the unique representative that is locally absolutely continuous and satisfies the integral formula, and step 5.1 shows that it is continuous; together these prove the "exactly one continuous representative" assertion, including $p=1$ in step 2.1 and $p=\mathrm{\infty }$ through steps 2.2, 3.1 and 4.1. Step 6.1 proves the finite-endpoint extension and its uniqueness. If the interval is unbounded, no endpoint extension is claimed, and if the interval is a single point, the interval is not open and the statement does not apply; for the zero class $u=0$ the representative $u^{\ast }=0$ is locally absolutely continuous, the formula reads $0-0=0$, and it is the unique one by step 4.1. The Axiom of Choice is used only through [F21], which supplies the Countable and Dependent Choice hypotheses of [F2], [F5], [F8], [F10] and [F11]; the intervals $J_n$ are explicitly parameterised and the only countable selections are the representatives $u_n$ of step 3.1, licensed by Countable Choice. $\mathrm{\square }$