A remark to the \(g\)-CLT result by Morgenthaler (Q2752172)

Let \(\lambda\) be the Lebesgue probability measure on \([0,1].\) Suppose a function \(g\) satisfies NEWLINE\[NEWLINE g(t)\geq 0 \quad \text{and}\quad \int^{1}_{0}g(t) dt=1. \tag{1}NEWLINE\]NEWLINE Define a probability measure \(\mu\) on \(\mathbb{R}\) by NEWLINE\[NEWLINE \mu(A)=\frac{1}{\lambda(s)}\int_{S}N_{0,g(t)}(A) dt, \quad S\subset [0,1].NEWLINE\]NEWLINE A sequence \(\{ \varphi_{n}\}\) of random variables on Lebesgue probability space is said to obey the \(g\)-CLT if the convergence in law NEWLINE\[NEWLINE \frac{1}{A_{n}}\sum^{n}_{k=1}a_{k}\varphi_{k}\rightarrow \frac{1}{\lambda(S)}\int_{S}N_{0,g(t)}(\cdot) dt \quad \text{as } n\rightarrow \infty. NEWLINE\]NEWLINE Morgenthaler proved that for any orthogonal sequence \(\{ \varphi_{n}\}\), there exists a subsequence \(\{ \varphi_{n_{k}}\}\) obeying the \(g\)-CLT for some bounded function \(g\) satisfying (1). Moreover he also proved the converse, that is, for any bounded function \(g\), there exists an orthogonal sequence obeying the \(g\)-CLT. In this paper the authors prove that the reverse holds without the boundedness condition on \(g.\) Also they investigate integrability of the sequence.NEWLINENEWLINEFor the entire collection see [Zbl 0964.00044].