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If $Z_n$ has probability at least p at $z_n$, with $p>0$ and $z_n\to z$, every weak limit has probability at least p at z. Apply the closed-set bound of the Portmanteau theorem to arbitrarily small closed intervals containing z and eventually $z_n$, then decrease their radii. Thus a persistent atom prevents convergence to a continuous law, including a chi-squared law. This provides a direct check when bounded-information estimation fails standard asymptotic-normal theory.

A weakly closed uniformly tight family of Borel probability measures on a Polish space is weakly compact. Embed the underlying space into a compact metric closure, take subsequential pushforward limits, and use the closed-set Portmanteau theorem on the common tightness compact sets. The limit retains full mass in the original space; inverse continuity of the probability pushforward embedding theorem then returns the weak limit.

For a Hadamard differentiable functional with continuous linear derivative $L$, define $R_t(h)=(\Phi (s+th)-\Phi (s))/t-L(h)$. For every compact $K$ and positive $ϵ$, some positive $\delta ,a$ make $|R_t(h)|<ϵ$ whenever $\mathrm{dist}(h,K)<\delta$ and $0<|t|<a$. If this failed, choose $t_j\to 0$, directions within $1/j$ of $K$ and remainders at least $ϵ$. Compactness gives a convergent subsequence of nearby points of $K$, hence convergent directions. The derivative definition and continuity of $L$ give remainder tending to zero, a contradiction. With a tight weak limit, the Portmanteau theorem makes random directions lie near a suitable compact set with arbitrarily large probability. Thus this lemma proves the functional delta method without assuming separability of the entire normed space or uniform differentiability on its whole unit ball.

Write the constant limit as $c$. First, convergence in distribution to a constant implies convergence in probability: the Portmanteau theorem applied to the closed set $\{y:|y-c|\ge \epsilon \}$ givesNow compare characteristic functions without imposing independence. For fixed $t\in \mathbb{R}$,The second bound uses $|e^{iu}-e^{iv}|\le \min \{2,|u-v|\}$. By convergence in probability, the limiting upper bound is $|t|\epsilon$; then let $\epsilon \downarrow 0$. Meanwhile $X_n$ has weak convergence of random variables to $X$, so $\mathbb{E}{e}^{it{X}_n}\to \mathbb{E}{e}^{itX}$. Thus the characteristic function of the sum tends to $e^{itc}\mathbb{E}{e}^{itX}$, the characteristic function of $X+c$, which is continuous at zero. The Lévy continuity theorem provesThis is the additive case of the Slutsky theorem, with its proof supplied. The constant limiting marginal removes the dependence obstruction in part (b).

Uniform integrability means ${\sup }_n\mathbb{E}[|X_n|1_{\{|X_n|>K\}}]\to 0$ as $K\to \infty$. In particular the first absolute moments are uniformly bounded. By the Portmanteau theorem, $\mathbb{E}|X|\le {\mathrm{liminf}}_n\mathbb{E}|X_n|<\infty$. Let $T_K(x)=\max (-K,\min (x,K))$, a bounded continuous truncation. Convergence in distribution gives $\mathbb{E}{T}_K(X_n)\to \mathbb{E}{T}_K(X)$. ThereforeThe first term tends to zero by uniform integrability, the second by integrability of $X$. Hence $\boxed{\mathbb{E}{X}_n\to \mathbb{E}X}$.

A real functional $\Phi$ is Hadamard differentiable at $s_0$ if there exists a continuous linear functional $L={\dot{\Phi }}_{s_0}:S\to \mathbb{R}$ such that, whenever $t_j\ne 0$, $t_j\to 0$ and $h_j\to h$ in the normed vector space,If $\Phi$ is defined only on a subset, require $s_0+t_j{h}_j$ to lie in its domain. The variation of the directions $h_j$ is essential: convergence along each fixed direction alone is a weaker differentiability condition. Tangential Hadamard differentiability restricts the limit directions to a specified subspace; here the derivative is defined on all of $S$.

We prove the functional delta method using compact sets, so do not need an unjustified application of the Skorokhod representation theorem on an arbitrary nonseparable normed vector space. As usual for function-space convergence in distribution, the limiting random element is tight, and the real statistics are measurable, or convergence is understood through outer expectations. These are automatic under the usual separable complete random-element formulation; the empirical-process formulation used below has a tight separably supported limit.

For $t\ne 0$ putHadamard differentiability says $R_{t_j}(h_j)\to 0$ for every convergent sequence of directions, since continuity gives $L(h_j)\to L(h)$. It implies a Hadamard remainder near a compact set: for every compact $K\subset S$ and $ϵ>0$ there are $a>0$ and $\delta >0$ such thatTo prove it, suppose it failed. Choose $t_j\to 0$ and $h_j$ at distance less than $1/j$ from $K$ with $|R_{t_j}(h_j)|\ge ϵ$. Choose $k_j\in K$ with $\Vert h_j-k_j\Vert <2/j$. A subsequence of $k_j$ converges by compactness, so the corresponding $h_j$ converge too. The defining derivative limit contradicts the lower bound on $R_{t_j}(h_j)$.

Now let $Z_n=r_n(X_n-s_0)\stackrel{d}{\to }X$ and $t_n=1/r_n$. Given $\eta >0$, tightness of a probability measure provides a compact $K$ with $\mathbb{P}(X\in K)>1-\eta$. For the open $\delta$-neighbourhood $K^{\delta }$, the Portmanteau theorem givesOn $K^{\delta }$ the preceding uniform remainder is less than $ϵ$ for all sufficiently large $n$. Consequently ${\mathrm{limsup}}_n\mathbb{P}(|R_{t_n}(Z_n)|\ge ϵ)\le \eta$. Since $\eta$ is arbitrary, this remainder tends to zero in probability. We have proved the actual expansionContinuity of $L$ and the continuous mapping theorem give $L(Z_n)\stackrel{d}{\to }L(X)$. The vanishing remainder and Slutsky theorem therefore yieldThis derivation uses the derivative definition, rather than assuming a uniform Fréchet remainder over every bounded set. If outer expectations are needed, the same compact-neighbourhood argument uses outer probabilities and gives the corresponding empirical-process functional delta method.

For the statistical application, use the sample's empirical distribution function and the plug-in estimatorHere the printed $L^{\infty }$ denotes bounded pointwise functions with the supremum norm, namely ${\ell }^{\infty }(\mathbb{R})$, rather than equivalence classes modulo Lebesgue null sets. Assume Hadamard differentiability at the true $F$ and the usual measurability of the scalar estimator. The Donsker theorem for empirical distribution functions and the proved functional delta method giveNo continuity of $F$ is needed for this application. Since the derivative is continuous and linear and the bridge is a centred Gaussian process, $Z_F$ is a centred Gaussian real random variable, possibly identically zero.

Finally, a finite real weak limit ensures the required boundedness in probability. Choose $M$ so large that $\mathbb{P}(|Z_F|\ge M)<ϵ/2$. Applying the closed-set part of the Portmanteau theorem to $\{z:|z|\ge M\}$ givesHence there is $n_0$ such that, for every $n\ge n_0$,In particular the upper-tail condition in the question holds. The plug-in estimator has estimation error $O_{\mathbb{P}}(n^{-1/2})$, even when its limiting variance is zero.

Let $g:M^{\prime }\to \mathbb{R}$ be bounded and continuous. The composition $h=g\circ f$ is bounded and measurable, and every discontinuity point of $h$ is a discontinuity point of $f$. ThusThe Portmanteau theorem includes the null-discontinuity criterion: if $X_n\stackrel{d}{\to }X$ and a bounded measurable function is continuous at $X$ almost surely, then its expectations converge. HenceSince this holds for every bounded continuous $g$, it proves the continuous mapping theorem conclusion $f(X_n)\stackrel{d}{\to }f(X)$.

A homeomorphic embedding of a metric space induces a homeomorphic embedding of its probability measures for their weak topologies. The inverse continuity follows from the open-set Portmanteau theorem, since every relatively open set is the trace of an ambient open set. This does not give surjectivity onto $P(Z)$ when points of $Z$ are outside the embedded image.