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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.