The integrand is positive and continuous on the interval in question. Substitute $t=\log (1+x)$ to obtainIf $\alpha \ge 1$, the exponential factor is at most one and the comparison integral of $t^{-2}$ converges. If $\alpha <1$, exponential growth dominates $t^2$, so the integrand eventually exceeds one and the integral diverges. ThusThe logarithmic square is what makes the boundary value $\alpha =1$ integrable.

Let $A_n=\{x:f(x)\ge 1/n\}$. To avoid presupposing finite measure of $A_n$, first truncate it: $A_{n,k}=A_n\cap [-k,k]$ is measurable and has finite measure. Its indicator divided by $n$ is integrable and is bounded above by $f$. The allowed monotonicity property givesThus every $A_{n,k}$ is null. Their countable union $A_n$ is null by part (a), and so is $E={\bigcup }_n{A}_n$. Every positive value of $f$ is at least $1/n$ for some integer $n$, hence $\{f>0\}=E$. Consequently $f=0$ outside a null set.

A null set has Lebesgue outer measure zero: for every $\epsilon >0$ it can be covered by countably many open intervals with total length below $\epsilon$. If $E_j$ are null, cover $E_j$ by intervals of total length below $\epsilon 2^{-j}$, for $j\ge 1$. The union of all these covers is countable and covers ${\bigcup }_j{E}_j$ with total length below $\epsilon$. As $\epsilon$ is arbitrary, the countable union is null.

Take the curl. A gradient has zero curl, and the supplied identity, with $\nabla \cdot u=0$ and $\nabla \cdot \omega =\nabla \cdot (\nabla \times u)=0$, givesTherefore $D\omega /Dt=(\omega \cdot \nabla )u$. In planar flow $\omega =(0,0,{\partial }_{x}v-{\partial }_{y}u)$ and ${\partial }_{z}u=0$, so the stretching term vanishes. ConsequentlyVorticity is carried unchanged with each fluid particle. It need not be spatially uniform or independent of time at a fixed spatial point.

For steady flow the given equation gives $\nabla H=-\rho \omega \times u$. Dotting with $u$ yields $u\cdot \nabla H=0$, so $H$ is constant along each streamline. It can have different values on different streamlines. Dotting with $\omega$ also makes it constant along vortex lines where they are defined.

If the flow is additionally irrotational, $\omega =0$ and hence $\nabla H=0$. Thus $H$ is one constant on each connected fluid region, the stronger global Bernoulli conclusion.

Choose $Z=(A,{\mathcal{T}}_A)$ and let $f$ be the identity function on $A$. The composite $i\circ f$ is continuous by part (i). The stipulated universal implication therefore makes $f:(A,{\mathcal{T}}_A)\to (A,{\mathcal{T}}^{\prime })$ continuous. Its inverse image of any ${\mathcal{T}}^{\prime }$-open set is that set itself, so it must lie in ${\mathcal{T}}_A$. Thus $\boxed{{\mathcal{T}}^{\prime }\subseteq {\mathcal{T}}_A}$.

If $A\cap U$ is subspace-open, thenThe right side is open in $Z$ by the assumed continuity of the composition. This holds for every subspace-open set, proving $f$ is continuous into the subspace topology.

Every subspace-open set has the form $A\cap U=i^{-1}(U)$ for some $U\in \mathcal{T}$. Continuity of $i:(A,{\mathcal{T}}^{\prime })\to (X,\mathcal{T})$ makes every such inverse image open in ${\mathcal{T}}^{\prime }$. Hence $\boxed{{\mathcal{T}}_A\subseteq {\mathcal{T}}^{\prime }}$.

For $U\in \mathcal{T}$, the inverse image under inclusion is $i^{-1}(U)=A\cap U$, which is open in the subspace topology by definition. The inverse-image criterion for continuity therefore proves the inclusion is continuous.

A metric space is sequentially compact if every sequence in it has a subsequence converging to a point of the same space.

A sequence converges if there is $x\in X$ such that, for every $ϵ>0$, some $N$ satisfies $d(x_n,x)<ϵ$ whenever $n\ge N$. The limit must belong to $X$, not merely to a larger ambient space.

The truncations $g_n=x^{1/2}{1}_{[1/n,1]}$ are measurable, nonnegative, and increase almost everywhere to $g$. As before their integrals agree with the corresponding Riemann integrals, givingMonotone convergence therefore gives the finite value $\boxed{\int g=2/3}$. Nonnegativity makes this the integral of the absolute value as well, proving Lebesgue integrability.

Use the nonnegative increasing truncations $f_n=x^{-2}{1}_{[1/n,1]}$. They increase almost everywhere to the specified function. On each compact interval the continuous function has equal Riemann and Lebesgue integrals, so the fundamental theorem of calculus givesMonotone convergence yields $\int f=+\infty$. Since a nonnegative function is Lebesgue integrable exactly when its integral is finite, this function is not Lebesgue integrable.

The monotone convergence theorem states that if measurable nonnegative functions satisfy $f_n\uparrow f$ almost everywhere on a measure space, thenallowing either side to be $+\infty$.

Let $f(x)=x^4-6x^3+11x^2-5x$. The node polynomial isIt vanishes at all four nodes, so subtracting it leaves the data unchanged. Since $f(x)-\omega (x)=x$, uniqueness of polynomial interpolation gives

The polynomial $p(x)=x^2-2x$ already has degree at most three and gives the specified data at every node. Uniqueness of polynomial interpolation therefore gives

The expected counts in the final two separate classes are too small for a reliable chi-squared asymptotic approximation. Pooling the last three classes gives the tail class $\ge 7$, with observed frequency $8+3+1=12$ and expected frequency $5.464+1.473+0.447=7.384$. This exceeds the usual expected-count guideline of 5. Pooling only the last two would leave an expected count of $1.920$, still too small. Pooling requires recomputing the contribution from the pooled totals, rather than adding the old separate contributions.

Fit the Poisson distribution using the estimated daily mean. For $k=0,\dots ,8$, its expected frequency isand $E_{\ge 9}=N\left(1-{\sum }_{k=0}^8{e}^{-\hat{\mu }}{\hat{\mu }}^k/k!\right)$. Each row's contribution to the Pearson chi-squared statistic is $(O_k-E_k{)}^2/E_k$. For example the zero-count class contributes $(162-126.784{)}^2/126.784\approx 9.781$. Summing the class contributions gives the test statistic; expected frequencies depend on the fitted model, while observed frequencies come directly from counting days.

For independent daily counts $X_1,\dots ,X_N$ modeled by a Poisson distribution, the log-likelihood isDifferentiating gives the maximum-likelihood estimator $\hat{\mu }=\overline{X}$. Here the observed frequencies sum to $N=1096$, and the expected frequencies are consistent with $\hat{\mu }\approx 2.157$ (their ratio $E_1/E_0$ estimates the same parameter). The final class is open-ended: its single observation is only reported as at least 9, so its exact contribution to the sample total cannot be recovered from this grouped table alone. The underlying ungrouped daily observations determine the exact sample mean.