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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).

Write $A_n=n^{-1}{\sum }_{j=1}^n{X}_j$ and $\phi (t)=\mathbb{E}{e}^{it{X}_1}$. The given two-sided limit $|f(u)|/u\to 0$ implies $|f(u)|/|u|\to 0$, so $f(u)=o(|u|)$. For each fixed real $t$, independence and the characteristic function of a sum of independent variables giveSet $z_n=iat/n+f(t/n)$. For $t\ne 0$, $n{z}_n\to iat$ and $z_n=O(n^{-1})$. The local Taylor expansion of the complex logarithm at $1$ givesThus $\mathbb{E}{e}^{it{A}_n}\to e^{iat}$; at $t=0$ the identity is immediate. The limit is continuous at zero and is the characteristic function of the constant random variable $a$. The Lévy continuity theorem proves the weak law from a characteristic-function expansion:No integrability hypothesis has been used. The conclusion is also convergence in probability by (d)'s convergence in distribution to a constant implies convergence in probability. If the word “constant” were to allow complex $a$, the characteristic function symmetry $\phi (-t)=\overline{\phi (t)}$ forces $a=\overline{a}$, so it is necessarily real.

Let $F_n$ be the distribution function of $X_n$. The limiting constant zero has distribution function $F(x)=0$ for $x<0$ and $F(x)=1$ for $x\ge 0$. For every $\epsilon >0$, both $-\epsilon$ and $\epsilon$ are continuity points of $F$. Thus convergence in distribution implies $F_n(-\epsilon )\to 0$ and $F_n(\epsilon )\to 1$. ThereforeThis provesThe same proof after subtracting a fixed real $c$ proves convergence in distribution to a constant implies convergence in probability. A deterministic limit fixes the coupling automatically; the counterexample in (c) exploits a nonconstant limit whose probability distribution alone does not fix that coupling.

If independent and identically distributed random variables have characteristic function $\phi (t)=1+iat+o(|t|)$ at zero, then their sample mean has convergence in distribution to $a$: its characteristic function is $\phi (t/n{)}^n\to e^{iat}$. This is also convergence in probability by convergence in distribution to a constant implies convergence in probability.