Write $A_n=n^{-1}\sum_{j=1}^nX_j$ and $\varphi(t)=\mathbb E e^{itX_1}$. The given two-sided limit $|f(u)|/u\to0$ implies $|f(u)|/|u|\to0$, so $f(u)=o(|u|)$. For each fixed real $t$, <independence> and the <characteristic function of a sum of independent variables> give
$$
\mathbb E e^{itA_n}=\left(1+\frac{iat}{n}+f(t/n)\right)^n.
$$
Set $z_n=iat/n+f(t/n)$. For $t\ne0$, $nz_n\to iat$ and $z_n=O(n^{-1})$. The local <Taylor expansion> of the <complex logarithm> at $1$ gives
$$
n\log(1+z_n)=nz_n+O(n|z_n|^2)\longrightarrow iat.
$$
Thus $\mathbb E e^{itA_n}\to e^{iat}$; 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>:
$$
\boxed{\frac{X_1+\cdots+X_n}{n}\xrightarrow{d}a.}
$$
No <integrable random variable>[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 $\varphi(-t)=\overline{\varphi(t)}$ forces $a=\overline a$, so it is necessarily real.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 201 / 6 / b / Solution

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

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 Incoming links: Weak law from a characteristic-function expansion