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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

Let $f:[a,b]\to \mathbb{R}$ be continuous. If it were unbounded, one could choose $x_n\in [a,b]$ with $|f(x_n)|>n$. The Bolzano-Weierstrass theorem gives a subsequence $x_{n_j}\to x\in [a,b]$, but continuity would then give $f(x_{n_j})\to f(x)$, contradicting unboundedness.

Now let $M=\sup f([a,b])$. Choose $x_n$ with $f(x_n)>M-1/n$. A convergent subsequence and continuity give $f(x)=M$ at its limit. Applying the same argument to $-f$ shows that the infimum is attained. This proves the extreme value theorem on a closed bounded interval.

The functionis continuous and bounded, but attains neither its infimum $0$ nor its supremum $1$. For the second example, enumerate the rationals in $[0,1]$ as $(q_n)$ and setEvery nondegenerate interval contains infinitely many rational numbers, hence some $q_n$ with arbitrarily large $n$; therefore $\psi$ is unbounded on every such interval.

For the running extrema, compactness lets us writeThe function $f$ is uniformly continuous on $[a,b]$. Given $\epsilon >0$, choose $\delta >0$ such that $|f(u)-f(v)|<\epsilon$ whenever $|u-v|<\delta$. If $x<y$ and $y-x<\delta$, every new value $f(\xi )$ with $x\le \xi \le y$ is at most $f(x)+\epsilon \le M(x)+\epsilon$. Since $M(y)\ge M(x)$,Interchanging $x,y$ handles the other order. Applying this argument to $-f$ proves continuity of $m$ as well.

Finally fix $T>0$ and put $h(x)=g(x+T)-g(x)$. For each positive integer $n$, there must be some $x_n>n$ with $|h(x_n)|<1/n$. Otherwise, for some $n$ the continuous function $h$ would satisfy $|h(x)|\ge 1/n$ on the interval $(n,\infty )$. By the intermediate value theorem, $h$ would have a constant sign there. The valueswould then increase or decrease by at least $1/n$ at every step, contradicting boundedness of $g$. Thus $x_n\to \infty$ and

A Lagrange top is an axially symmetric rigid body with one point fixed, centre of mass a distance $l$ along its symmetry axis, principal moments $I_1=I_2,I_3$, mass $M$, and Euler angles $\theta ,\phi ,\psi$ in gravity $g$.

The cyclic coordinates give $p_{\psi }=I_3(\dot{\psi }+\dot{\phi }\cos \theta )$ and $p_{\phi }=I_1\dot{\phi }{\sin }^2\theta +p_{\psi }\cos \theta$; the energy $E$ is the third integral. Eliminating $\dot{\phi },\dot{\psi }$ from $E$ and setting $u=\cos \theta$, with $\dot{u}=-\dot{\theta }\sin \theta$, yields ${\dot{u}}^2=f(u)$ after multiplication by $1-u^2$; all denominators cancel and $f$ is cubic.

This is Poisson regression with independent counts and log mean $\log {\mu }_d=1.5624+0.1658d$. The residual deviance $9.90$ on $12$ degrees of freedom has $p=0.625$, and the plotted curve tracks the data, so there is no evidence of poor fit.

The fitted doubling time is $\log 2/0.1658=4.16$ days. A 95% interval for the growth coefficient is $0.1658\pm 1.96(0.0166)=(0.133,0.198)$, corresponding to doubling times $(3.50,5.21)$ days. Seven days, whose rate is about $0.099$, is unsupported.