Absolute Continuity and the Sharp Fundamental Theorem of Calculus

Fix $s<t$ and $\epsilon >0$. Absolute continuity gives $\delta >0$ such that every permitted finite interval family of total length below $\delta$ has total $F$-increment below $\epsilon \mathrm{/}2$. Put $D=\{x\in (s,t):F^{\mathrm{\prime }}(x)=0\}$ and $\eta =\epsilon \mathrm{/}(2(t-s+1))$. At every $x\in D$, differentiability supplies arbitrarily short closed intervals $I=[u,v]\subseteq [s,t]$ centred at $x$ such that $\mathrm{\mid }F(v)-F(u)\mathrm{\mid }\le \mathrm{\mid }F(v)-F(x)\mathrm{\mid }+\mathrm{\mid }F(x)-F(u)\mathrm{\mid }<\eta (v-u)\mathrm{.}$ These intervals form a fine cover of $D$.

Apply The Vitali covering theorem for fine covers on the real line with residual outer measure below $\delta$. It gives pairwise disjoint $I_1,\dots ,I_N$ from the fine cover such that ${\lambda }^{\ast }\text{\hspace{0.17em}⁣}(D\setminus {\bigcup }_{j=1}^N{I}_j)<\delta \mathrm{.}$ If $N_0=(s,t)\setminus D$, then ${\lambda }^{\ast }(N_0)=0$, so $R=[s,t]\setminus {\bigcup }_{j=1}^N{I}_j$ has outer measure below $\delta$ (the endpoints add no outer measure).

Order the selected intervals from left to right. The finitely many closed gaps between them, including the end gaps from $s$ and to $t$, have pairwise disjoint open interiors and total length $\lambda (R)<\delta$. Absolute continuity therefore bounds the sum of the $F$-increments over the gaps by $\epsilon \mathrm{/}2$. The estimate in step 1.1 bounds the corresponding sum over the selected intervals by $\eta {\sum }_{j=1}^N\mathrm{\mid }{I}_j\mathrm{\mid }\le \eta (t-s)<\epsilon \mathrm{/}2.$

Telescoping across the alternating selected intervals and gaps gives $\mathrm{\mid }F(t)-F(s)\mathrm{\mid }<\epsilon$. Since $\epsilon$ is arbitrary, $F(t)=F(s)$. This includes $a=b$, and arbitrary $s,t$ prove constancy.

Given $\epsilon >0$, apply Absolute continuity of the integral to $\mathrm{\mid }f\mathrm{\mid }$ and obtain $\delta >0$.

For disjoint intervals $[u_j,v_j]$ of total length below $\delta$, their union $E$ has measure below $\delta$, and ${\sum }_j\mathrm{\mid }{I}_f(v_j)-I_f(u_j)\mathrm{\mid }\le {\int }_E\mathrm{\mid }f\mathrm{\mid }<\epsilon$.

This is the defining finite-family condition for absolute continuity; the empty family and $a=b$ give zero sums.

The hierarchy $C^1$ implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation makes $F$ continuous and BV. By Jordan decomposition for functions of bounded variation, write $F=P-N$ with $P,N$ nondecreasing.

The monotone derivative theorem For a nondecreasing function, the derivative is measurable and integrable and its integral is bounded by the total increase gives derivatives $P^{\mathrm{\prime }},N^{\mathrm{\prime }}$ almost everywhere and $P^{\mathrm{\prime }},N^{\mathrm{\prime }}\in L^1$. Hence $F^{\mathrm{\prime }}=P^{\mathrm{\prime }}-N^{\mathrm{\prime }}$ exists almost everywhere and is integrable.

Degenerate intervals have no interior derivative assertion and the zero function in $L^1$, as required.

Assume (1). By Absolutely continuous functions have integrable derivatives, put $f=F^{\mathrm{\prime }}\in L^1$ a.e.; the first $L^1$ FTC The indefinite integral of an $L^1$ function is differentiable almost everywhere says $I_f^{\mathrm{\prime }}=f$ a.e.

The corollary The indefinite integral of an $L^1$ function is absolutely continuous makes $I_f$ AC, so $H:=F-F(a)-I_f$ is AC and has derivative zero a.e. The zero-derivative theorem makes $H$ constant; $H(a)=0$, giving (2).

Conversely, (2) says $F=F(a)+I_{F^{\mathrm{\prime }}}$, and the same corollary makes it AC. The singleton interval satisfies both clauses directly.

Given $\epsilon >0$, choose $\delta$ from Absolute continuity on a compact interval. The endpoints have a finite image, so it suffices to consider $E\cap (a,b)$. Outer regularity Assuming countable choice, the Lebesgue outer measure of an arbitrary subset of ${\mathbb{R}}^n$ is the infimum of the measures of the open sets containing it gives an open $U\subseteq (a,b)$ containing it with $\lambda (U)<\delta$.

Write $U={⨆}_j(a_j,b_j)$. On each $[a_j,b_j]$, continuity gives points $c_j,d_j$ of minimum and maximum value, ordered so that $c_j\le d_j$. Hence $F([a_j,b_j])$ is an interval of length $\mathrm{\mid }F(d_j)-F(c_j)\mathrm{\mid }$. The intervals $[c_j,d_j]$ are disjoint and have total length at most $\lambda (U)$; the AC estimate applies to every finite subfamily, so its nonnegative countable sum is at most $\epsilon$.

Subadditivity gives ${\lambda }^{\ast }(F(E\cap (a,b)))\le {\sum }_j\mathrm{\mid }F(d_j)-F(c_j)\mathrm{\mid }\le \epsilon$. Letting $\epsilon \downarrow 0$ and restoring the finite endpoint image proves property $(N)$ as defined in Luzin's property $(N)$ on a compact interval.

Fix $\eta >0$. On the differentiability set, split $E$ into the levels $(k-1)\eta \le \mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }<k\eta$ and then into sets on which the differentiability estimate has one common radius. A cover of each latter set by intervals shorter than that radius shows that its image has outer measure at most $k\eta$ times the outer measure of the set: two points in one covering interval have image distance at most $k\eta$ times its length.

Sum the level estimates. Since $(k-1)\eta \le \mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }$ on level $k$, this gives ${\lambda }^{\ast }(F(E))\le {\int }_E\mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }+\eta \lambda (E)$. Letting $\eta \downarrow 0$ proves the first assertion; the integrability convention is that of Integrable real and complex functions, and their integrals.

If $[a,b]\setminus D$ is null, property $(N)$ Luzin's property $(N)$ on a compact interval makes $F(E\setminus D)$ null. Subadditivity and step 2.1 applied to $E\cap D$ give the stated consequence. The singleton interval is immediate.

If $F$ is AC, its defining interval estimate directly gives continuity and bounded variation; it has $(N)$ by Absolutely continuous functions have Luzin's property $(N)$.

Conversely assume continuity, BV, and $(N)$. The normalized Jordan decomposition Jordan decomposition for functions of bounded variation gives $F=F(a)+P-N$, and For a nondecreasing function, the derivative is measurable and integrable and its integral is bounded by the total increase gives $F^{\mathrm{\prime }}=P^{\mathrm{\prime }}-N^{\mathrm{\prime }}\in L^1$ wherever both derivatives exist. The BV differentiability theorem Every function of bounded variation is differentiable almost everywhere makes that a full-measure set.

Apply the consequence of Luzin's property $(N)$ gives an integral growth estimate to each interval $[u_j,v_j]$ in a finite disjoint family. Since continuity makes $F([u_j,v_j])$ an interval, $\mathrm{\mid }F(v_j)-F(u_j)\mathrm{\mid }\le {\int }_{u_j}^{v_j}\mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }$. The absolute continuity of the integral Absolute continuity of the integral now gives the finite-family AC condition.

Thus the reverse implication holds; both directions include the singleton interval.

Let $Z$ be the countable exceptional set. It is null by Every at most countable subset of $\mathbb{R}$ has measure zero. If $A\subseteq [a,b]$ is null, Luzin's property $(N)$ gives an integral growth estimate applied to $A\setminus Z$ gives ${\lambda }^{\ast }(F(A\setminus Z))=0$, while $F(A\cap Z)$ is countable and hence null. Thus $F$ has property $(N)$.

For $u<v$, continuity gives the interval between $F(u)$ and $F(v)$ as a subset of $F([u,v])$. The latter image is contained in $F([u,v]\setminus Z)\cup F([u,v]\cap Z)$, whose second part is countable and whose first part has outer measure at most ${\int }_u^v\mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }$ by the growth lemma. Hence $\mathrm{\mid }F(v)-F(u)\mathrm{\mid }\le {\int }_u^v\mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }$. Summing over any partition bounds its variation by ${\int }_a^b\mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }<\mathrm{\infty }$, so $F$ has bounded variation.

Since $F$ is continuous by hypothesis, step 1.1 gives property $(N)$ and step 2.1 gives bounded variation. The reverse direction of Banach--Zarecki characterisation of absolute continuity therefore makes $F$ AC, and Fundamental theorem of calculus for absolutely continuous functions gives the displayed reconstruction formula at every endpoint.

Absolute continuity Absolute continuity on a compact interval implies continuity, hence $\mathrm{\mid }F\mathrm{\mid }\le M$ and $\mathrm{\mid }G\mathrm{\mid }\le N$ on the compact interval. Choose the two AC moduli with error $\epsilon \mathrm{/}(2\max \{M,1\})$ and $\epsilon \mathrm{/}(2\max \{N,1\})$.

For every interval, $\mathrm{\mid }F(v)G(v)-F(u)G(u)\mathrm{\mid }\le M\mathrm{\mid }G(v)-G(u)\mathrm{\mid }+N\mathrm{\mid }F(v)-F(u)\mathrm{\mid }$. Summing over a sufficiently short disjoint family gives $<\epsilon$.

This is the defining condition, also when a factor or the interval is degenerate.

The product lemma makes $FG$ AC. At the common full-measure set where $F^{\mathrm{\prime }}$ and $G^{\mathrm{\prime }}$ exist, the ordinary product rule gives $(FG{)}^{\mathrm{\prime }}=F^{\mathrm{\prime }}G+F{G}^{\mathrm{\prime }}$. The sharp FTC makes $F^{\mathrm{\prime }},G^{\mathrm{\prime }}$ integrable, and continuity on the compact interval bounds $F,G$, so both products are integrable.

Apply Fundamental theorem of calculus for absolutely continuous functions to $FG$ and integrate the displayed derivative.

Rearranging gives the stated identity, with identical zero sides on $[a,a]$.

First take $f={\mathbf{1}}_{(r,s)}$. The clipped function $(g-r{)}^{+}-(g-s{)}^{+}$ is AC; its derivative is ${\mathbf{1}}_{(r,s)}(g)g^{\mathrm{\prime }}$ almost everywhere (on a level set of an AC increasing function, $g^{\mathrm{\prime }}=0$ almost everywhere). The sharp FTC evaluates its integral as the length of $(r,s)\cap [g(a),g(b)]$.

The interval-indicator identity in step 1.1 extends first to the algebra of finite unions of intervals and then, by The monotone class generated by an algebra equals the sigma-algebra it generates, to all Borel indicators. If $N$ is Lebesgue null, Assuming countable choice, the Lebesgue outer measure of an arbitrary subset of ${\mathbb{R}}^n$ is the infimum of the measures of the open sets containing it covers it by open sets $O$ of arbitrarily small length. The open-set case then makes $g^{-1}(N)\cap \{g^{\mathrm{\prime }}>0\}$ null: on $\{g^{\mathrm{\prime }}\ge 1\mathrm{/}k\}$ its outer measure is at most $k{\int }_{g^{-1}(O)}{g}^{\mathrm{\prime }}\le k\lambda (O)$, and take the union over $k$. Every Lebesgue-measurable $f$ has a Borel representative off a null set, so this observation makes $f(g)g^{\mathrm{\prime }}$ agree almost everywhere with a measurable weighted composition. Simple-function approximation and Monotone convergence for the integral now extend the identity to every nonnegative measurable $f$.

Apply step 2.1 to $f^{+}$ and $f^{-}$. Applying it also to $\mathrm{\mid }f\mathrm{\mid }$ gives ${\int }_a^b\mathrm{\mid }f(g(x))\mathrm{\mid }{g}^{\mathrm{\prime }}(x)dx={\int }_{g(a)}^{g(b)}\mathrm{\mid }f(y)\mathrm{\mid }dy<\mathrm{\infty }$, so $f(g)g^{\mathrm{\prime }}\in L^1$ and subtraction gives the displayed formula. Constant and singleton cases have both integrals zero. This is also the conclusion of the cited Heil corollary.

If $a=b$, both conclusions are vacuous almost-everywhere assertions on a null interval. If $c=d$, then $g$ and $F\circ g$ are constant and the product is zero almost everywhere. Hence suppose $a<b$ and $c<d$.

By The indefinite integral of an $L^1$ function is absolutely continuous, $F$ is AC, and The indefinite integral of an $L^1$ function is differentiable almost everywhere gives $F^{\mathrm{\prime }}=f$ almost everywhere. Since $F$ is real-valued, Absolutely continuous functions have Luzin's property $(N)$ gives its image-null property $(N)$. The real-valued functions $g$ and $H:=F\circ g$ are AC, so Fundamental theorem of calculus for absolutely continuous functions makes both differentiable almost everywhere.

Heil's cited chain-rule theorem applies: $g$, $F$, and $H$ are differentiable almost everywhere, and $F$ maps null sets to null sets. Its conclusion holds for every function $h=F^{\mathrm{\prime }}$ almost everywhere. Taking $h=f$ yields $H^{\mathrm{\prime }}=(f\circ g)g^{\mathrm{\prime }}$ almost everywhere, including on the pullback of the exceptional set for $F^{\mathrm{\prime }}$. No assertion that this pullback is null is needed.

Since the real-valued function $H$ is AC, its derivative is integrable by Fundamental theorem of calculus for absolutely continuous functions. The product is finite-valued under our convention and equals this measurable derivative outside a null set by step 3.1. Completeness of Lebesgue measure therefore makes the product measurable, and almost-everywhere equality gives $f(g)g^{\mathrm{\prime }}\in L^1[a,b]$.

The lemma identifies $f(g)g^{\mathrm{\prime }}$ with $(F\circ g{)}^{\mathrm{\prime }}$ almost everywhere and proves it integrable.

Apply Fundamental theorem of calculus for absolutely continuous functions to the AC function $F\circ g$.

Its endpoint formula is exactly the displayed equality, including $a=b$.

The sharp FTC gives $F(v)-F(u)={\int }_u^v{F}^{\mathrm{\prime }}$, so every partition sum is at most ${\int }_a^x\mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }$.

Approximate the sign of $F^{\mathrm{\prime }}$ by a finite step function and use its sign-change endpoints as a partition; the corresponding partition sum approaches ${\int }_a^x\mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }$. Thus the supremum defining Total-variation function on a compact interval equals that integral.

Since $\mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }\in L^1[a,b]$, The indefinite integral of an $L^1$ function is absolutely continuous makes the right side an absolutely continuous function, and The indefinite integral of an $L^1$ function is differentiable almost everywhere gives its derivative $\mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }$ almost everywhere. Step 2.1 identifies that function with $V_F$.

If $F$ is $L$-Lipschitz, every difference quotient is bounded by $L$; wherever the derivative exists, $\mathrm{\mid }{F}^{\mathrm{\prime }}\mathrm{\mid }\le L$. Absolute continuity itself follows from $C^1$ implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation.

Conversely, the sharp FTC gives $F(y)-F(x)={\int }_x^y{F}^{\mathrm{\prime }}$, whose absolute value is at most $L\mathrm{\mid }y-x\mathrm{\mid }$.

This proves both directions; it includes $L=0$ and $a=b$.

For $\epsilon >0$, use the AC modulus of $F$ from Absolute continuity on a compact interval with tolerance $\epsilon \mathrm{/}L$ when $L>0$.

Sum $\mathrm{\mid }h(F(v_j))-h(F(u_j))\mathrm{\mid }\le L\mathrm{\mid }F(v_j)-F(u_j)\mathrm{\mid }$ over a short disjoint family.

If $L=0$ the composite is constant; otherwise step 2.1 is the AC condition.

Choose the AC modulus $\eta$ of $h$ from Absolute continuity on a compact interval and set $\delta =\eta \mathrm{/}L$ for $L>0$.

Images under monotone $g$ of a disjoint ordered interval family have disjoint interiors, and their total length is at most $L$ times the original total length.

Apply the AC condition for $h$ to these image intervals. If $L=0$, $g$ and the composite are constant.

The derivative of $g$ is bounded on $(0,1]$ and $g^{\mathrm{\prime }}(0)=0$, so If $f$ is continuous on an interval $I$ and $\mathrm{\mid }{f}^{\mathrm{\prime }}\mathrm{\mid }\le M$ at every interior point, then $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }\le M\mathrm{\mid }x-y\mathrm{\mid }$ for all $x,y\in I$, so $f$ is Lipschitz with constant $M$ and uniformly continuous on $I$ and $C^1$ implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation show that $g$ is absolutely continuous. Since $h^{\mathrm{\prime }}(y)=1\mathrm{/}(2\sqrt{y})$ is integrable and $h(y)-h(0)={\int }_0^y{h}^{\mathrm{\prime }}(t)dt$, Fundamental theorem of calculus for absolutely continuous functions shows that $h$ is absolutely continuous.

For $x>0$, $(h\circ g)(x)=\mathrm{\mid }x\sin (1\mathrm{/}x)\mathrm{\mid }$. On the alternating half-waves accumulating at $0$, its total variation has a positive contribution comparable to $1\mathrm{/}n$, and the harmonic series diverges. Thus $h\circ g$ is not of bounded variation.

Every absolutely continuous function has bounded variation by $C^1$ implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation, so $h\circ g$ is not absolutely continuous. Hence the universal composition assertion is false.

The function $C$ is continuous and nondecreasing, hence has bounded variation, and $C(K)=[0,1]$.

It maps the null Cantor set onto $[0,1]$, so it fails property $(N)$.

Absolute continuity would imply $(N)$ by Absolutely continuous functions have Luzin's property $(N)$. Therefore $C$ is a continuous BV function that is not AC, refuting the statement.

The function $C$ is continuous, satisfies $C(0)=0$ and $C(1)=1$, and has derivative $0$ almost everywhere; thus $C^{\mathrm{\prime }}\in L^1[0,1]$.

Its endpoint increment is $1$, whereas ${\int }_0^{1}0=0$.

Thus the asserted every-$x$ reconstruction formula fails without an additional hypothesis such as AC.

The derivative of $g$ is bounded on $(0,1]$ and $g^{\mathrm{\prime }}(0)=0$, so $g$ is Lipschitz and AC. Also $h^{\mathrm{\prime }}(y)=1\mathrm{/}(2\sqrt{y})\in L^1[0,1]$, so $h$ is AC.

For $x>0$, $(h\circ g)(x)=\mathrm{\mid }x\sin (1\mathrm{/}x)\mathrm{\mid }$. Its successive alternating half-waves contribute a quantity comparable to $1\mathrm{/}n$ to total variation, so the variation diverges.

Every AC function on a compact interval has bounded variation by $C^1$ implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation. Hence $h\circ g$ is not AC, and the statement is false.

Since $F^{\mathrm{\prime }}(x)=1\mathrm{/}(2\sqrt{x})$ belongs to $L^1[0,1]$, $F(x)={\int }_0^x{F}^{\mathrm{\prime }}$ is absolutely continuous.

If $F$ were $L$-Lipschitz, then $\sqrt{x}\le Lx$ for $x>0$, hence $1\mathrm{/}\sqrt{x}\le L$, which fails as $x\downarrow 0$.

Therefore $F$ is AC but not Lipschitz.

The function $F$ is $C^1$ on every interval $[1\mathrm{/}(n+1),1\mathrm{/}n]$, hence Lipschitz there. These intervals and the singleton $\{0\}$ form a countable cover, so $F$ maps every null set to a null set and has property $(N)$ as in Luzin's property $(N)$ on a compact interval.

Since $F^{\mathrm{\prime }}(x)=\sin (1\mathrm{/}x)-\cos (1\mathrm{/}x)\mathrm{/}x$ for $x>0$, alternating subintervals give infinite variation. Thus $F$ is not BV.

Every absolutely continuous function on a compact interval is BV by $C^1$ implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation. Hence this property-$(N)$ function is not AC.