# Absolute continuity with respect to a Rajchman measure

A measure $\sigma$ on $\mathbb{T}$ (the unit circle in the complex plain) is called a Rajchman measure if $\hat{\sigma}(n)\rightarrow0$ as $|n| \rightarrow \infty$.

I want to prove that if $\sigma\ll\mu$ and $\mu$ is a Rajchman measure on $\mathbb{T}$, then $\sigma$ is also a Rajchman measure.

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By Radon-Nykodym theorem, we can find a function $f$, integrable, such that $f\mu=\sigma$. This function can be approached in $L^1(\mu)$ by trigonometric polynomials, that is, functions of the form $$\sum_{j=-N}^Na_je^{2i\pi jx}.$$ Let $P_n$ a sequence of trigonometric polynomials such that $\lVert f-P_n\rVert_{L^1(\mu)}\leq 1/n$. Then $$|\widehat\sigma(n)|=\left|\int_{\Bbb T}e^{inx}\sigma(dx)\right|=\left|\int_{\Bbb T}e^{inx}f(x)\mu(dx)\right|\leq \frac 1N+\left|\int_{\Bbb T}e^{inx}P_N(x)\mu(dx)\right|.$$ Say $P_N(x)$ has the form $\sum_{j=-K_N}^{K_N}a_{N,j}e^{ijx}$. Then $$\left|\int_{\Bbb T}e^{inx}P_N(x)\mu(dx)\right|\leq\sum_{j=-K_N}^{K_N}|a_{n,j}||\widehat \mu(n+j)|.$$ We get $$|\widehat\sigma(n)|\leq \frac 1N+\sum_{j=-K_N}^{K_N}|a_{n,j}||\widehat \mu(n+j)|,$$ and taking the $\limsup$, $|n|\to +\infty$, we deduce that for each $N$, $$\limsup_{|n|\to +\infty}|\widehat\sigma(n)|\leq\frac 1N,$$ hence $\mu$ is Rajchman.