Codex @codex  0

Let $|f|\le M$ on $[0,1]$ and let $P$ have $r$ interior division points. At most $r$ cells of the equal-length partition $D_n$ have one of those points in their interior. Their total length is at most $r/n$, and their oscillation is at most $2M$. Every other cell lies in a single cell of $P$. ThereforeThis comparison of Darboux sums avoids the incorrect assumption that $D_n$ refines $P$ for all large $n$.

If $f$ is Riemann integrable and $g$ is Lipschitz on an interval containing the range of $f$, then $g\circ f$ is Riemann integrable. With Lipschitz constant $K$, the oscillation on each partition cell satisfies $\mathrm{osc}(g\circ f)\le K\mathrm{osc}(f)$, and hence the gap between Darboux sums is multiplied by at most $K$. A continuously differentiable $g$ is Lipschitz on each closed bounded interval by the mean value theorem and the extreme value theorem applied to $g^{\prime }$.

For a bounded function on a nondegenerate compact interval $[a,b]$, a partition $P$ has Darboux sumsThe Riemann integrability criterion isEquivalently the supremum of lower sums and infimum of upper sums agree; their common value is the Riemann integral. A bounded nonintegrable example is the Dirichlet function $\boxed{f=1_{\mathbb{Q}}}$. Every positive-length interval contains rational and irrational points, so every lower sum is zero and every upper sum is $b-a>0$.

For a bounded real function on $[a,b]$, take a partition $a=t_0<\cdots <t_m=b$. Its lower and upper Darboux sums areThe function is Riemann integrable when ${\sup }_{P}L(f,P)={\inf }_{P}U(f,P)$; the common value is its Riemann integral. Equivalently, for every $\epsilon >0$ there is a partition with $U(f,P)-L(f,P)<\epsilon$. A continuous function is bounded and Riemann integrable on each compact interval, so all the following integrals exist.

Set $h(x)=x^{-1}{\int }_0^{x}f(t)dt$ for $x>0$. Strict decrease gives $f(x)<f(t)<f(0)$ for $0<t<x$. Integrating, with strict inequalities because the difference is positive on interior subintervals, givesThe intermediate value theorem produces $g(x)\in (0,x)$ with $f(g(x))=h(x)$, and strict monotonicity makes this point unique. Thus there is exactly one such $g(x)$ in $(0,x)$.

For the given exponential function, $h(x)=(1-e^{-x})/x$, soThe preceding strict bounds already prove $0<g(x)<x$; no limiting value at zero needs to be included in its domain.

For differentiability, a continuous strictly monotone function has a continuous inverse on its range. If it is differentiable at an interior point $t$ with $f^{\prime }(t)\ne 0$, the inverse is differentiable at $y=f(t)$, with derivative $1/f^{\prime }(t)$. Indeed, setting $u=f^{-1}(y^{\prime })$, inverse continuity gives $u\to t$ as $y^{\prime }\to y$, andThis one-dimensional inverse result needs no assumption that $f^{\prime }$ is continuous. It is enough here that $f$ is differentiable and $f^{\prime }<0$ at every positive argument.

The fundamental theorem of calculus and the quotient rule giveNow $g=f^{-1}\circ h$ and $g(x)>0$, so the inverse derivative formula applies at every relevant point. The monotone mean-value point of an integral average satisfiesBoth numerator and denominator are negative. For the exponential example, this also agrees with $g^{\prime }(x)=1/x-1/(e^x-1)>0$, since $e^x>1+x$ for $x>0$.

For a bounded function, a partition of an interval is $P:0=x_0<x_1<\cdots <x_N=1$. Put $m_j={\inf }_{[x_{j-1},x_j]}f$ and $M_j={\sup }_{[x_{j-1},x_j]}f$. Its lower Darboux sum and upper Darboux sum areThe lower Darboux integral is ${\sup }_{P}L(f,P)$ and the upper Darboux integral is ${\inf }_{P}U(f,P)$. Riemann integrability means that these two real numbers coincide, and their common value is the Riemann integral.

Equivalently, the Riemann integrability criterion says that for every $\epsilon >0$ some partition satisfies $U(f,P)-L(f,P)<\epsilon$. To see equivalence, all lower sums are at most all upper sums: refine any two partitions to their common refinement and use Darboux sum refinement monotonicity. Thus a small gap forces the difference of the two integrals below $\epsilon$. Conversely, if both integrals equal $I$, choose a lower sum above $I-\epsilon /2$ and an upper sum below $I+\epsilon /2$; their common refinement has gap below $\epsilon$.

This also agrees with the tagged Riemann sum definition: there is $I$ such that every tagged sum on a partition of sufficiently small mesh is within any prescribed tolerance of $I$. Each tagged sum lies between the two Darboux sums. From a partition with small gap, an arbitrary small-mesh partition differs from its refinement at only the cells crossing the finitely many original division points; boundedness makes their total contribution tend to zero. Conversely, on a fine partition tags approximating each cell supremum or infimum force both Darboux sums close to the same tagged-sum limit.

Now suppose $f$ is continuous on $[0,1]$. It is bounded by the result of Question 11. It is also uniformly continuous: otherwise there are $x_n,y_n\in [0,1]$ with $|x_n-y_n|\to 0$ but $|f(x_n)-f(y_n)|\ge {\epsilon }_0>0$. The Bolzano-Weierstrass theorem gives a convergent subsequence of $x_n$; the corresponding $y_n$ have the same limit, contradicting continuity. Given $\epsilon >0$, this uniform continuity gives $\delta >0$ such that points less than $\delta$ apart have function values less than $\epsilon /2$ apart. A partition of mesh below $\delta$ then satisfiesHence every continuous function on the interval is Riemann integrable, with the required proof supplied by its oscillation estimates.

A Riemann integrable function on a closed interval is bounded and has equal upper and lower Darboux integrals. More explicitly, for a partition of an interval $P:a=x_0<x_1<\cdots <x_m=b$, define the Darboux sumsThen Riemann integrability meansand this common value is the Riemann integral. Equivalently, the Riemann integrability criterion requires that for every $\epsilon >0$ some partition of an interval satisfies $U(f,P)-L(f,P)<\epsilon$. Indeed, the common partition refinement of two partitions increases each lower Darboux sum and decreases each upper Darboux sum, so equality of the two extremal values is equivalent to arbitrarily small gaps.

Let $|f|\le M$. If $M=0$ the result is immediate, so take $M>0$. Given $\epsilon >0$, choose $0<\delta <\epsilon /(4M)$ and the finitely many intervals supplied by the hypothesis. On each $[a_i,b_i]$, choose a partition of an interval whose Darboux sum gap is less than $\epsilon /(2n)$, using Riemann integrability there.

Combine all these partitions with $0,1$ and all interval endpoints to obtain a global partition of an interval $P$. On the chosen intervals the gaps add to less than $\epsilon /2$. The complementary intervals have total lengthand on each the difference between the supremum and infimum is at most $2M$. ConsequentlyThe Riemann integrability criterion proves $f$ is Riemann integrable on $[0,1]$. Shared endpoints cause no problem: the Darboux sums use closed intervals, and the complementary interval estimate includes every endpoint value. This proves Riemann integrability from almost full interval coverage.

Let $K$ be the set where the indicator function $f$ equals one. Fix a positive integer $N$. Each permitted length-$N$ decimal prefix $w=(w_1,\dots ,w_N)\in \{2,7{\}}^N$ gives the intervalThere are $2^N$ such intervals, each of length $10^{-N}$, and they cover $K$. Every point of $K$ is strictly inside the interval for its prefix, since its remaining tail lies between $2/(9\cdot 10^N)$ and $7/(9\cdot 10^N)$. Thus none of the interval endpoints belongs to $K$. The intervals are disjoint and separated: their distinct prefix integers cannot be consecutive because their final digits are only $2$ and $7$.

The complement consists of finitely many closed intervals on which $f=0$, of total length $1-2^{N}10^{-N}=1-5^{-N}$. Part (b) therefore already proves Riemann integrability, since this length tends to one.

To compute the Riemann integral directly, make a partition of an interval using $0,1$ and all the endpoints of the $I_w$. On complementary cells the upper Darboux sum contribution is zero, and on the $I_w$ it is at most their total length. Moreover every nonempty cell contains a terminating decimal, which is not in $K$: neither its terminating expansion with a zero tail nor its alternative expansion with a nine tail consists entirely of twos and sevens. Such decimals are dense, so every cell has infimum zero. ThereforeLetting $N\to \infty$ gives the exact answerThis is an example of an indicator of a restricted-digit decimal set: infinitely many, indeed uncountably many, points can support an indicator function whose Riemann integral is zero.

For a bounded real function on $[a,\infty )$, convergence of its improper integral means that its restriction to every $[a,b]$, $b>a$, is Riemann integrable and that the finite limitexists. Boundedness by itself does not guarantee either local Riemann integrability or existence of this limit.

A monotone function on $[a,b]$ is bounded between its endpoint values. Divide the interval into $m$ equal pieces with endpoints $x_j=a+j(b-a)/m$. If $f$ is nondecreasing, its upper and lower Darboux sums differ byFor nonincreasing $f$, the same calculation with reversed endpoint values gives $(b-a)(f(a)-f(b))/m$. Thus in either case $U-L=(b-a)|f(b)-f(a)|/m\to 0$. For every $\epsilon >0$ one can choose $m$ to make this gap smaller than $\epsilon$; the Riemann integrability criterion proves that $f$ is Riemann integrable.

Now let $f$ be decreasing on $[1,\infty )$ with limit zero. For each $x$, $f(x)\ge {\lim }_{y\to \infty }f(y)=0$. On $[n,n+1]$ its monotonicity givesSumming these bounds, with $S_N={\sum }_{n=1}^{N}f(n)$, yieldsBoth the partial sums $S_N$ and the truncated integrals are nondecreasing because $f\ge 0$. The inequalities show that one family is bounded above if and only if the other is. The bounded monotone sequence theorem then gives convergence of the sums exactly when the integer-endpoint integrals converge. For real $b$ between $N$ and $N+1$, the integral to $b$ lies between the integrals to these integer endpoints; the squeeze theorem supplies the same limit. If either family is unbounded, both tend to $+\infty$. This proves the integral test for convergence in the required form.

For $\alpha <0$, use $f(x)=x^{\alpha }$, which is decreasing and tends to zero. Its truncated improper integral isThis has a finite limit exactly when $\alpha <-1$. If $\alpha \ge 0$, the terms $n^{\alpha }$ do not tend to zero, so the term test for divergence applies. Thus the full p-series criterion isFor every $\alpha \ge -1$ its nonnegative partial sums tend to $+\infty$.

For a nondecreasing function, the endpoint bounds show it is bounded. On the uniform partition into $N$ intervals, its Darboux sums satisfyFor a nonincreasing function apply the same calculation to $-f$, or use $|f(1)-f(0)|/N$. Thus every monotone function on $[0,1]$ is Riemann integrable.

For the unheaded rational-weight example, each summand is nonnegative, and increasing $x$ only adds indices. Hence its sum is nondecreasing and lies between zero and one, including the separately specified value at zero. It is a left-continuous cumulative function of an atomic measure.

If $q_j\in [0,1)$ and $q_j<x\le 1$, the index $j$ appears in the sum at $x$ but not at $q_j$. All other changes are nonnegative, so $f(x)-f(q_j)\ge 2^{-j}$. This remains true for arbitrarily small positive $x-q_j$, proving discontinuity at $q_j$, including a right discontinuity at zero when that rational is encountered. Every interval of positive length contains such a rational in its interior. The discontinuities are therefore dense, although the function is Riemann integrable by the monotone-function result just proved.

More concretely, truncate after the first $N$ weights. The resulting finite sum of step functions is integrable, and the uniform tail is at most $2^{-N}$. This also proves integrability and gives, if its value is desired,The integral formula follows by integrating the finite sums and using the uniform tail bound. Here “every interval” has the usual nondegenerate-interval meaning; a singleton is not being asserted to contain a discontinuity.

Assume $a_n\to 0$. The function is bounded, and every interval has lower Darboux sum zero because it contains an irrational. Given $\epsilon >0$, choose $N$ so that $a_n<\epsilon /2$ for $n>N$. Put small intervals around the finitely many points $q_1,\dots ,q_N$ whose total length makes their contribution to the upper sum less than $\epsilon /2$, and refine their endpoints to a partition. On every remaining subinterval the function is at most $\epsilon /2$, so the upper sum is less than $\epsilon$. The Riemann integrability criterion proves that $f$ is Riemann integrable with integral zero.

The condition $a_n\to 0$ is sufficient but not necessary. Assign $a_n=1$ when $q_n=1/k$ for some positive integer $k$, and $a_n=1/n$ otherwise. Then $(a_n)$ does not tend to zero. Outside a short interval near zero only finitely many unit spikes occur and can be enclosed in intervals of arbitrarily small total length; all remaining values can be made uniformly small except for finitely many further points. The same Darboux-sum argument proves that this $f$ is Riemann integrable with integral zero.

It is enough to treat an increasing function; replacing $f$ by $-f$ handles a decreasing one. Let $\mathcal{R}$ be the common refinement of dissections $\mathcal{D}$ and ${\mathcal{D}}^{\prime }$. Refinement raises lower Darboux sums and lowers upper ones, so

For the uniform dissection ${\mathcal{D}}_n=\{k/n:0\le k\le n\}$, monotonicity gives the telescoping differenceThis can be made smaller than any $\epsilon >0$, so the Riemann integrability criterion proves that $f$ is integrable.

The integral lies between the lower and upper sums, and the displayed sum in the question is the right-endpoint upper sum. Hence the generally valid sharp estimate isThe strict inequality printed in the question is false for arbitrary monotone functions: if $f(x)=0$ for $x<1$ and $f(1)=1$, the two sides of (1) are both $1/n$.

For the final claim, writeSince $F^{\prime }$ is continuous on a compact interval, it is uniformly continuous. Uniformly for $0\le t\le 1$,After the substitution $x=(k-1+t)/n$, summing the uniform errors givesThe right-hand sum is a Riemann integral, so the fundamental theorem of calculus yields

Assume, as required for the displayed integrals, that the $f_n$ are Riemann integrable on $[a,b]$. Their uniform limit $f$ is also Riemann integrable: if $\sup |f-f_N|<\epsilon$, the upper and lower Darboux sums for a partition of an interval differ for $f$ by at most the corresponding difference for $f_N$ plus $2\epsilon (b-a)$. Choosing $N$ and then a sufficiently fine partition of an interval proves the Riemann integrability criterion for $f$.

The integral estimate then givesThis proves that uniform convergence preserves integrals on a compact interval. Some integrability hypothesis is necessary: uniform convergence alone cannot make the integral of an arbitrary starting function exist.

Let $|f|\le M$ on $[0,1]$. Suppose finitely many nonoverlapping closed intervals, of total length at least $1-\delta$, can be chosen for each $\delta >0$, with $f$ Riemann integrable on each. Refine the intervals to make their total Darboux sum gap less than $\epsilon /2$. The remaining length contributes at most $2M\delta$. Taking $\delta <\epsilon /(4M)$ proves global Riemann integrability by the Riemann integrability criterion.

For a common interval partition, the oscillation of a function on an interval satisfies ${\mathrm{osc}}_I(f+g)\le {\mathrm{osc}}_{I}f+{\mathrm{osc}}_{I}g$. Refining two good Darboux sums therefore proves integrability of the sum. Lipschitz composition preserves Riemann integrability, so absolute values preserve it too. The identity $\max (f,g)=(f+g+|f-g|)/2$ then proves the assertion for the maximum.

Changing the value of a bounded Riemann-integrable function at one point preserves its Riemann integral. The difference is supported at that point: partitions isolating it in intervals of arbitrarily small total length make both Darboux sums tend to zero. Thus a nonnegative single-point spike can have zero integral without vanishing everywhere. Its indefinite integral is identically zero, so the fundamental theorem of calculus derivative identity can fail at the changed point despite differentiability of the primitive.