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Let $\varphi (u)=\int e^{iuy}\mu (dy)$. Since $1-\cos (uy)\ge 0$, the Tonelli theorem giveswith the fraction defined to be one at zero. For $|x|\ge 1$, $\sin x/x\le \sin 1$. To verify the constant, the function is even; on $[1,\pi ]$ it decreases because $(x\cos x-\sin x{)}^{\prime }=-x\sin x<0$ on $(0,\pi )$ and $x\cos x-\sin x$ starts from zero. On $[\pi ,\infty )$, $\sin x/x\le 1/\pi <\sin 1$. Consequently the integrand is at least $1-\sin 1$ on $|y|\ge \lambda$, proving the characteristic-function tail boundFor pointwise convergence ${\varphi }_n\to \varphi$, where $\varphi$ is a characteristic function, the dominated convergence theorem applies on each finite integration interval, since the integrands lie between zero and two. HenceContinuity of $\varphi$ at zero and $\varphi (0)=1$ make the right side tend to zero as $\lambda \to \infty$. Given $\epsilon >0$, choose $\lambda$ so it is below $\epsilon /2$. The tail probability is then below $\epsilon$ for all sufficiently large $n$. Enlarge the compact interval to control the finitely many remaining measures, each of which has vanishing tails. This proves uniform tightness of the entire sequence, not just of its tail. It is the characteristic-function tightness bound principle with the sharper displayed constant.

For the integrable mean-zero summands, differentiation at zero is justified byThe dominated convergence theorem gives ${\varphi }^{\prime }(0)=i\mathbb{E}{X}_1=0$. DefineIt is continuous away from zero by continuity of the characteristic function, and continuous at zero by the zero derivative. Since $|\varphi (u)|\le 1$, the finite geometric series yields, for $n\ge 1$,and the case $n=0$ has both sides zero. Thus the requested uniform-in-$n$ estimate isThis is the characteristic-function derivative under an absolute first moment estimate.

Independence makes the characteristic function of $S_n/n=(X_1+\cdots +X_n)/n$ equal to $\varphi (u/n{)}^n$. Therefore $|1-\varphi (u/n{)}^n|\le |u|\psi (u/n)$. Applying the already proved characteristic-function tail bound to its distribution, for every $\epsilon >0$,The convergence follows because ${\sup }_{0\le u\le 1/\epsilon }\psi (u/n)\to 0$, and the remaining integral of $u$ is finite. Hence $\boxed{S_n/n\to 0\text{ in probability}}$, the mean-zero weak law of large numbers. This deduction uses the tail bound directly and does not presuppose the Lévy continuity theorem.

On the real line, Prohorov's theorem states that a sequence of probability measures is uniformly tight if and only if every subsequence has a further subsequence converging weakly to a probability measure. Here uniformly tight means that, for every $\epsilon >0$, some compact interval contains mass at least $1-\epsilon$ for every measure in the sequence. This is relative sequential compactness for weak convergence of probability measures, not merely the existence of one convergent subsequence.

We first prove the characteristic-function tail bound. Put $c_0=1-\sin 1>0$. For $y\ne 0$,At $y=0$ the expression is zero by continuity. If $|y|\ge \lambda$, the supplied elementary bound, with $t=|y|/\lambda$, makes it at least $c_0$. The integrand is nonnegative, so Tonelli's theorem yieldsThus one valid universal constant isThe estimate controls spatial tails using only the characteristic function near zero.

Apply it to each ${\mu }_n$. For fixed $\lambda$, pointwise convergence of the characteristic functions and the bound $|{\varphi }_n|\le 1$ allow dominated convergence on the finite integration interval, givingContinuity of a characteristic function and $\varphi (0)=1$ make the last expression tend to zero as $\lambda \to \infty$. Given $\epsilon >0$, choose $\lambda$ making it less than $\epsilon /2$; then for all sufficiently large $n$, the preceding tail is less than $\epsilon$. Enlarge the compact interval to handle the finitely many remaining probability measures. This proves uniform tightness of the whole sequence.

By Prohorov's theorem, every subsequence has a further weakly convergent subsequence, say with limit $\nu$. The real and imaginary parts of $e^{iuy}$ are bounded continuous functions, so along that subsequenceThe uniqueness theorem for characteristic functions implies $\nu =\mu$. All subsequential weak limits are therefore $\mu$. If the original sequence failed to converge weakly to $\mu$, a bounded continuous test function and a subsequence would keep its integrals a fixed distance from the limiting integral; a further weakly convergent subsequence would contradict that. HenceThis supplies the required proof of the Lévy continuity theorem in the stated case.

For the final request, the printed $Q$ is not defined; the intended function is $\varphi$, the characteristic function of a summand. We prove that ${\varphi }^{\prime }(0)=ia$ without assuming integrability. Write $Y_n=S_n/n$. Independence gives $\mathbb{E}{e}^{iu{Y}_n}=\varphi (u/n{)}^n$. Moreover convergence in probability to $a$ implies uniform convergence of these characteristic functions on each bounded frequency interval: for $|u|\le T$ and $\delta >0$,First let $n\to \infty$, then $\delta \downarrow 0$.

The potential logarithm-branch ambiguity needs to be resolved, not discarded. By continuity of $\varphi$ near zero, its principal complex logarithm is defined at $u/n$ for all $|u|\le 1$ and all large $n$, with value zero at $u=0$. Alsouniformly on $[-1,1]$, so $\mathrm{Log}{\Psi }_n(u)\to 0$ uniformly there. The two continuous logarithms differ by an integer multiple of $2\pi i$:That integer is continuous in $u$ and zero at zero, hence identically zero. ConsequentlyFor arbitrary nonzero $h\to 0$, take $n=\lfloor 1/|h|\rfloor$ and $u=nh$. Eventually $1/2\le |u|\le 1$, so dividing the preceding error by $u$ shows $\mathrm{Log}\varphi (h)/h\to ia$. Exponentiating at zero proves the derivative forced by the weak law:This is the weak law forces a characteristic-function derivative argument. Its estimates use bounded exponentials, not $\mathbb{E}|X_1|$. If $X_1$ is integrable, dominated convergence permits differentiation inside the expectation and gives ${\varphi }^{\prime }(0)=i\mathbb{E}{X}_1$, so then $a=\mathbb{E}{X}_1$.

Absolute integrability is genuinely unnecessary. For an explicit weak law without an absolute first moment, take a symmetric variable withwith mass $1-1/e$ at zero and no mass with $0<|X|<e$. This is a valid law: the specified tail decreases from $1/e$ to zero, and its derivative defines the continuous part beyond $e$. Its absolute first moment is infinite because ${\int }_e^{\infty }dx/(x\log x)=\infty$. For independent copies, set $Y_{n,j}=X_j{1}_{\{|X_j|\le n\}}$. Symmetry gives $\mathbb{E}{Y}_{n,j}=0$, and integration of the tail givesFor the last estimate, split the integral at $\sqrt{n}$; the initial part is $O(\sqrt{n})$ and the remainder is at most $2n/\log n$, apart from a fixed initial constant. Chebyshev's inequality now gives $n^{-1}{\sum }_{j\le n}{Y}_{n,j}\to 0$ in probability. The probability that any truncation changes a summand is at most $n\mathbb{P}(|X|>n)=1/\log n\to 0$, so $S_n/n\to 0$ in probability although $\mathbb{E}|X|=\infty$. The derivative is zero, but the usual expectation of $X$ does not exist.

If either the positive or negative part is integrable, the finite weak-law limit does force full integrability. For example, if $\mathbb{E}{X}^{-}<\infty$, then $X\wedge A$ is integrable and its sample averages have the weak law of large numbers limit $\mathbb{E}(X\wedge A)$. Since they are bounded above by the original averages, $\mathbb{E}(X\wedge A)\le a$ for every $A$. Letting $A\to \infty$ shows $\mathbb{E}{X}^{+}<\infty$. The symmetric argument handles the other one-sided hypothesis. The nonintegrable example necessarily has both one-sided expectations infinite.