\documentstyle{SibMatJ} %\TestXML \topmatter \Author Song \Initial Y. \Gender he \Sign Yanbo Song \Email syb888202106\@163.com \AffilRef 1 \endAuthor \Affil 1 \Division School of Information Engineering \Organization Xi'an University \City Xi'an \Country China \endAffil \datesubmitted August 10, 2025\enddatesubmitted %\daterevised August 23, 2025\enddaterevised \dateaccepted August 23, 2025\enddateaccepted \UDclass ??? %\?11N05, 11N35, 11L07 \endUDclass \title On Primes of the Form $\lfloor\alpha p+\beta\rfloor$ in Piatetski\hbox{-}Shapiro %\? the, много без артикля Sequence \endtitle \abstract In this paper, we investigate whether there are infinitely many primes of the form $\lfloor\alpha p+\beta\rfloor$ in Piatetski\hbox{-}Shapiro sequence for almost all irrationals $\alpha>0$ (in the sense of Lebesgue measure). Our result provides an affirmative answer to this question by utilizing elementary and analytic methods, especially the `Partition of Unity' %\? and the estimations of exponential sums. \endabstract \keywords Goldbach conjecture, twin prime conjecture, Selberg's sieve method, Piatetski\hbox{-}Shapiro primes, partition of unity \endkeywords \endtopmatter \head 1. Introduction \endhead %\?начало слово в слово из предыдущей статьи, поэтому оттуда списала In 2009, motivated by the Goldbach conjecture and the twin prime conjecture, H. Li and H. Pan [1] proposed the following conjecture. \proclaim{Conjecture 1} Let $\alpha>0$ be an irrational, and let $\beta$ be a real. Then there exist infinitely many primes $p$ such that $\lfloor\alpha p+\beta\rfloor$ is also a prime. \endproclaim In connection with the above conjecture, Li and Pan proved the following weaker theorem. %Concerned with the above conjecture, H. Li and H. Pan [1] proved a weaker theorem as follows: \proclaim{Theorem 1 \rm[1,~Theorem~1]} Let $\beta$ be a real. Then for almost all irrational $\alpha>0$ {\rm(}in the sense of Lebesgue measure{\rm)}, $$ \limsup\limits_{x\rightarrow\infty}\frac{\pi_{\alpha,\beta}(x)(\log^{2}x)}{x}\geq 1, $$ where $$ \pi_{\alpha,\beta}(x)=\bigl\{p\leq x: \text{ both } p \text{ and } \lfloor\alpha p+\beta\rfloor \text{ are primes}\bigr\}. $$ \endproclaim On the other hand, the Piatetski\hbox{-}Shapiro sequence is a fundamental object in analytic number theory, named after the Soviet mathematician Israel Moiseevich Piatetski\hbox{-}Shapiro. It refers to sequences of the form $\{\lfloor n^c\rfloor\}_{n=1}^{\infty}$, where $c>1$ is a real. The theory of Piatetski\hbox{-}Shapiro prime %\?the без артикля всюду, наверное plays an important role in analytic number theory. In 1953 Piatetski\hbox{-}Shapiro [2] gives an asymptotic formula for the counting function of Piatetski\hbox{-}Shapiro prime for $c\in(1,12/11)$. It is worth emphasizing that the range $c\in(1,12/11)$ was improved by Kolesnik [3], Heath-Brown [4], Kolesnik [5], and Liu and Rivat~[6]. The best known result in this direction is due to Rivat and Sargos [7], which give an asymptotic formula for the counting function of Piatetski\hbox{-}Shapiro prime for $c\in(1,2817/2426)$. We remark that Rivat and Wu~[8] give the lower bound for the counting function for $c\in(1,243/205)$. This result is the best regarding the infinitude of the Piatetski\hbox{-}Shapiro primes. It is worth noting that some authors generalize the Piatetski\hbox{-}Shapiro primes, interested readers can see [9--12] for examples. %\? Considering the above discussions of the two lines of research, we proposed the following conjecture: \proclaim{Conjecture 2} Let $\alpha>0$ be an irrational, and let $\beta$ be a real. Then there exist infinitely many primes $p$ such that $\lfloor\alpha p+\beta\rfloor$ are primes in Piatetski\hbox{-}Shapiro sequence. \endproclaim We remark %\?note est' Remark that Guo [13] was the first to consider Piatetski\hbox{-}Shapiro primes in a Beatty sequence, Conjecture 2 can be seen as the generalization of Guo's work [13]. Concerned with the above conjecture, we proved a weaker theorem as follows: \proclaim{Theorem 2} Let $\alpha$, $\beta$, and $\frac{12}{13}<\gamma<1$ be reals, and let $P=\{\lfloor n^{\frac{1}{\gamma}}\rfloor:n\in \bold{N}\}$. Then for almost all irrational $\alpha>0$ {\rm(}in the sense of Lebesgue measure{\rm)}, $$ \limsup\limits_{x\rightarrow\infty}\frac{\pi^{\gamma*}_{\alpha,\beta}(x)(\log^{2}x)}{x^{\gamma}}\geq 1, $$ where $$ \pi^{\gamma*}_{\alpha,\beta}(x)=\{p\leq x: p \text{ is a prime},\ [\alpha p+\beta]\in P \text{ is a prime}\}. $$ \endproclaim \demo{Remark} In this remark, %\? we summarize the proof strategy of \Par*{Theorem 2} as follows. Following %\? Li and Pan [1], the approach is to reduce the problem to a sieve upper bound $\pi^{\gamma*}_{\alpha,\beta}(x)\ll x^{\gamma}/\alpha\log^2x$ (\Par*{Lemma~7}). In our case, the first additional complexity arises from the formula for the characteristic function of Piatetski\hbox{-}Shapiro primes (as presented in \Par*{Lemma 2}), while the second lies in estimating the exponential sum involving floor functions. To address these challenges, we employ the `Partition of Unity' %\? technique introduced in \Par*{Lemma 4}. \enddemo \demo{Some Notations} %\?может, убрать Let $p$ always denote a prime. Let $\mes(A)$ denote the Lebesgue measure of $A$. We will frequently use $\varepsilon$ with a slight abuse of notation to mean a small positive, % number, possibly a~different one each time. Given a real $t$, we write $e(t)=e^{2\pi it}$, $\{\alpha\}$ for the fractional part of $\alpha$, and $\lfloor\alpha\rfloor$ for the largest integer not exceeding $\alpha$. We recall that for functions $F$ and real nonnegative $G$ the notations $F\ll G$ and $F=O(G)$ are equivalent to the statement that the inequality $|F|\leq\alpha G$ holds for some constant $\alpha>0$. We also write $F\sim G$ to indicate that $F\ll G$ and $G\ll F$. \enddemo \head 2. Auxiliary Results \endhead For readers' convenience, we list some auxiliary results before proving \Par*{Theorem 2}. Firstly, we need the following approximation of Vaaler [14]. \proclaim{Lemma 1} Let $\psi(x)=x-\lfloor x\rfloor-1/2$. Then for any $H\geq1$ there are numbers $a_{h}$ and $b_{h}$ such that $$ \biggl|\psi(t)-\sum\limits_{0<|h|\leq H}a_{h}e(th)\biggr|\leq\sum\limits_{|h|\leq H}b_{h}e(th), $$ where $$ a_{h}\ll\frac{1}{|h|},\quad b_{h}\ll\frac{1}{H}. $$ \endproclaim We need the following lemma, which is Lemma 4 in [15]. \proclaim{Lemma 2} A natural $m$ has the form $\lfloor n^{c}\rfloor$ if and only if $\Cal{X}(m)=1$, where $$ \Cal{X}(m)=\gamma m^{\gamma-1}+\psi(-m^{\gamma})-\psi(-(m+1)^{\gamma})+O(m^{\gamma-2}), $$ and $\gamma=1/c$. \endproclaim \proclaim{Lemma 3} Let $f$ be three times continuously differentiable on a subinterval $I$ of $(N,2N]$. Suppose that for some $\lambda>0$, the inequalities $$ \lambda\ll|f^{'''}(x)|\ll\lambda $$ hold, where the implied constants are independent of $f$ and $\lambda$. Then $$ \sum\limits_{n\in I}e(f(n))\ll N\lambda^{1/6}+N^{3/4}+N^{1/4}\lambda^{-1/4}. $$ \endproclaim \demo{Proof} This is Theorem 2.6 in [16]. \qed\enddemo The following lemma is the so-called `Partition of Unity.' %\? \proclaim{Lemma 4} For $Z\in\bold{N}$, there exist periodic functions $g_z(t)$ $(0\leq z\leq Z)$ such that $$ \align &0b_{1}>0$, $12/13<\gamma<1$, $\alpha$, and $\beta$ are arbitrary reals. For each $0<\varepsilon<1/16800$ and all sufficiently large real $x$, there exists an exceptional set $J_{E}\subseteq(b_{1},b_{2})$ with $\mes(J_{E})=O(x^{-\varepsilon})$ such that for any square-free $d\leq x^{\frac{13\gamma}{12}-1}$ and irrational $\alpha\in(b_{1},b_{2})\diagdown J_{E}$, $$ |\{1\leq n\leq x:d|n\lfloor\alpha n+\beta\rfloor,\ \lfloor\alpha n+\beta\rfloor=\lfloor m^{\frac{1}{\gamma}}\rfloor\}|=\frac{\alpha^{\gamma-1}x^{\gamma}}{d}\prod\limits_{p\mid d}\Bigl(2-\frac{1}{p}\Bigr)+ O\Bigl(\frac{x^{\gamma-\varepsilon}}{d}\Bigr). $$ \endproclaim \demo{Proof} Let $$ S:=\sum\Sb\lfloor\alpha n+\beta\rfloor=\lfloor m^{1/\gamma}\rfloor \\ d|n\lfloor\alpha n+\beta\rfloor \endSb 1, \tag1 $$ and $$ E_{\alpha,\beta,d}(n)= \cases 1, &\text{if } d\mid \lfloor\alpha n+\beta\rfloor; \\ 0, &\text{otherwise}. \endcases $$ By \Par*{Lemma 2}, we have $$ \Cal{X}_{\gamma}(m)=\gamma m^{\gamma-1}+\psi(-m^{\gamma})-\psi(-(m+1)^{\gamma})+O(m^{\gamma-2}). \tag2 $$ Let $$ P=\bigl\{1\leq n\leq x:d\,|\,n %\? \lfloor\alpha n+\beta\rfloor,\ \lfloor\alpha n+\beta\rfloor=\lfloor m^{\frac{1}{\gamma}}\rfloor\bigr\}. \tag3 $$ So by \Tag(1) and \Tag(3), we have $$ \aligned S &=|P| \\ &=\sum\limits_{s|d}\ \bigl|\{1\leq n\leq x/s:\lfloor\alpha sn+\beta\rfloor\equiv0(\modulo d/s),\ (n,d/s)=1,\ ns\in P\}\bigr|\\ &=\sum\limits_{s|d}\sum\limits_{t|(d/s)}\mu(t)\bigl| \{1\leq n\leq x/s:\lfloor\alpha sn+\beta\rfloor\equiv0(\modulo d/s),\ t|n,\ ns\in P\}\bigr|\\ &=\sum\limits_{s|d,\,t|s}\mu(t)\bigl|\{1\leq n\leq x/s:\lfloor\alpha sn+\beta\rfloor\equiv0 (\modulo td/s),\ ns\in P\}\bigr|\\ &=\sum\limits_{s|d,\,t|s}\mu(t) \sum\Sb n\leq x/s\\td/s|\lfloor\alpha ns+\beta\rfloor\endSb (\gamma \lfloor\alpha ns+\beta\rfloor^{\gamma-1}+\psi(-\lfloor\alpha ns+\beta\rfloor^{\gamma})-\psi(-(\lfloor\alpha ns+\beta\rfloor+1)^{\gamma})\\ &\qquad+O(\lfloor\alpha ns+\beta\rfloor^{\gamma-2})) \\ &=S_1+S_2+O(x^{\gamma-1+\epsilon}), %\?\varepsilon всюду \endaligned \tag4 $$ where $$ \align S_1 &=\sum\limits_{s|d,t|s}\mu(t) \sum\Sb n\leq x/s\\td/s|\lfloor\alpha ns+\beta\rfloor\endSb \gamma \lfloor\alpha sn+\beta\rfloor^{\gamma-1},\\ S_2 &=\sum\limits_{s|d,t|s}\mu(t) \sum\Sb n\leq x/s\\td/s|\lfloor\alpha ns+\beta\rfloor\endSb (\psi(-\lfloor\alpha sn+\beta\rfloor^{\gamma})-\psi(-(\lfloor\alpha sn+\beta\rfloor+1)^{\gamma})). \endalign $$ By Lemma 7 in [1], we have for almost all $\alpha>0$ $$ \sum\limits_{n\leq x/s}E_{\alpha,\beta,td/s}(ns) =\frac{x}{td}(1+O(x^{-2\epsilon})). \tag5 $$ For $S_1$, by summation by parts we have $$ S_1=\frac{\alpha^{\gamma-1}x^{\gamma}}{d}\prod\limits_{p\mid d}\Bigl(2-\frac{1}{p}\Bigr)+O\Bigl(\frac{x^{\gamma-\varepsilon}}{d}\Bigr). \tag6 $$ For $S_2$, it is enough to prove $$ S^{'}_2:= \sum\Sb n\leq x/s\\td/s|\lfloor\alpha ns+\beta\rfloor\endSb (\psi(-\lfloor\alpha sn+\beta\rfloor^{\gamma})-\psi(-(\lfloor\alpha sn+\beta\rfloor+1)^{\gamma}))\ll\frac{x^{\gamma-\epsilon}}{d}. \tag7 $$ For $S'_2$, by \Par*{Lemma 5}, we have $$ \aligned S^{'}_2 &=\sum\limits_{n\leq x/s}E_{\alpha,\beta,td/s}(ns)(\psi(-\lfloor\alpha sn+\beta\rfloor^{\gamma})-\psi(-(\lfloor\alpha sn+\beta\rfloor+1)^{\gamma})) \\ &=\sum\limits_{n\leq x/s}(\psi(-\lfloor\alpha sn+\beta\rfloor^{\gamma})-\psi(-(\lfloor\alpha sn+\beta\rfloor+1)^{\gamma}))\frac{1}{td/s}\sum\limits_{a=1}^{dt/s}e\Bigl(\frac{as\lfloor\alpha sn+\beta\rfloor}{td}\Bigr). \endaligned \tag8 $$ By \Par*{Lemma 1} with $H=dx^{1-\gamma+\epsilon}$, we have $$ \align \psi(- &\lfloor\alpha sn+\beta\rfloor^{\gamma})-\psi(-(\lfloor\alpha sn+\beta\rfloor+1)^{\gamma})\\ &=\sum\limits_{0<\mid h\mid\leq H}a_{h}(e(h\lfloor\alpha sn+\beta\rfloor^{\gamma})-e(h(\lfloor\alpha sn+\beta\rfloor+1)^{\gamma}))\\ &\qquad+O\biggl(\ \sum\limits_{\mid h\mid \leq H}b_{h}(e(h\lfloor\alpha sn+\beta\rfloor^{\gamma}) +e(h(\lfloor\alpha sn+\beta\rfloor+1)^{\gamma}))\biggr). \endalign $$ So we have $$ \aligned S^{'}_2 &\ll \frac{s}{td}\sum\limits_{a=1}^{dt/s}\sum\limits_{0<\mid h\mid\leq H}\frac{1}{H} \biggl|\sum\limits_{n\leq x/s}(e(h\lfloor\alpha sn+\beta\rfloor^{\gamma})\biggr| \\ &\qquad+\frac{s}{td}\sum\limits_{a=1}^{dt/s}\sum\limits_{0<\mid h\mid\leq H}\frac{1}{h}\biggl|\sum\limits_{n\leq x/s}e\Bigl(\frac{as\lfloor\alpha sn+\beta\rfloor}{td}+h\lfloor\alpha sn+\beta\rfloor^{\gamma}\Bigr)\biggr| \\ &:=T_1+T_2+O\Bigl(\frac{x^{\gamma-\varepsilon}}{d}\Bigr). \endaligned \tag9 $$ We only treat $T_2$, it is enough to prove $$ \sum\limits_{n\leq x/s}e\Bigl(\frac{as\lfloor\alpha sn+\beta\rfloor}{td}+h\lfloor\alpha sn+\beta\rfloor^{\gamma}\Bigr)\ll\frac{x^{\gamma-\epsilon}}{d}. $$ By \Par*{Lemma 4} with $Z=d^2x^{2-2\gamma+\epsilon}$, we have $$ \align \sum\limits_{n\leq x/s}e &\Bigl(\frac{as\lfloor\alpha sn+\beta\rfloor}{td}+h\lfloor\alpha sn+\beta\rfloor^{\gamma}\Bigr) \\ &=\sum\limits_{n\leq x/s}e\Bigl(\frac{as\lfloor\alpha sn+\beta\rfloor}{td}+h\lfloor\alpha sn+\beta\rfloor^{\gamma}\Bigr)\sum\limits_{z=0}^{2Z-1}g_z(\alpha sn+\beta) \\ &=\sum\limits_{z=0}^{2Z-1}\sum\limits_{n\leq x/s}\Bigl(e\Bigl(\frac{as^2\alpha n+as\beta}{td}-\frac{asz}{2tdZ}+h\Bigl(\alpha sn+\beta-\frac{z}{2Z}\Bigr)^{\gamma}\Bigr)+O\Bigl(\frac{x^{\gamma-1}h}{Z}\Bigr)\Bigr) %\times g_z(\alpha sn+\beta) \\ &=\sum\limits_{z=0}^{2Z-1}\sum\limits_{n\leq x/s}\Bigl(e\Bigl(\frac{\alpha as^2n+as\beta}{td}+h\Bigl(\alpha sn+\beta-\frac{z}{2Z}\Bigr)^{\gamma}\Bigr)\Bigr)g_z(\alpha sn+\beta) \\ &\qquad+O\Bigl(\frac{x^{\gamma-\varepsilon}}{d}\Bigr) \\ \ll&\sum\limits_{z=0}^{2Z-1}\sum\limits_{|r|\leq Z(\log x)^4}|\beta_r^{(z)}|\biggl|\sum\limits_{n\leq x/s}\Bigl(e\Bigl(\frac{\alpha as^2n+as\beta}{td}+h\Bigl(\alpha sn+\beta-\frac{z}{2Z}\Bigr)^{\gamma}+r(\alpha sn+\beta)\Bigr)\Bigr)\biggr| \\ &\qquad+O\Bigl(\frac{x^{\gamma-\varepsilon}}{d}\Bigr). \endalign $$ For the innermost sum, in the interval $[1,\frac{x^{\gamma-\epsilon}}{d}]$ we estimate the innermost sum trivially, and next we divide the interval $[\frac{x^{\gamma-\epsilon}}{d},\frac{x}{s}]$ of the innermost sum into intervals of the form $[\frac{x/s}{2^{j+1}},\frac{x/s}{2^j}]$ and apply \Par*{Lemma~4} to the innermost sum, we have $$ \aligned \sum\limits_{n\leq x/s}e &\Bigl(\frac{as\lfloor\alpha sn+\beta\rfloor}{td}+h\lfloor\alpha sn+\beta\rfloor^{\gamma}\Bigr) \\ &\ll Z(x^{1/2+\gamma/6}h^{1/6}d^{1/2}+x^{3/4}+x^{1-\gamma/4}h^{-1/4})\log x+O\Bigl(\frac{x^{\gamma-\varepsilon}}{d}\Bigr) \\ &\ll\frac{x^{\gamma-\varepsilon}}{d}. \endaligned \tag10 $$ By \Tag(7), \Tag(9), and \Tag(10), we have $$ S_2\ll \frac{x^{\gamma-\varepsilon}}{d}, $$ this together with \Tag(6) completes the proof of this lemma. %\?2 this \qed\enddemo \proclaim{Lemma 7} Suppose that $b_{2}>b_{1}>0$, $12/13<\frac{1}{c}=\gamma<1$, and $\alpha$, $\beta$ are arbitrary reals. And we denote $P$ %\?by the set $\{p: \lfloor n^{\frac{1}{\gamma}}\rfloor=\lfloor\alpha p+\beta\rfloor, \ p \text{ is a prime}\}$. For all sufficiently large real $x$, there exists an exceptional set $J_{E}\subseteq(b_{1},b_{2})$ with $\mes(J_{E})=O(x^{-\varepsilon})$ such that for irrational $\alpha\in(b_{1},b_{2})\diagdown J_{E}$, we have $$ |\{1\leq p\leq x,\ p\in P: p \text{ is a prime}\}|\ll\frac{x^{\gamma}}{\log^{2} x}. $$ \endproclaim \demo{Proof} Let $d$ be a square-free number %\? such that $d\leq x^{\frac{13\gamma}{12}-1}$. Let $z=d^{1/2}$, $$ P(z)=\prod\limits_{pb_1>0$. Let $$ \Cal{F}=\Big\{\alpha\in (b_1,b_2): \limsup\limits_{x\rightarrow\infty}\pi^{\gamma*}_{\alpha,\beta}(x)\log^2x/x^{\gamma}<1\Big\}, $$ and $$ \Cal{F}_n= \Big\{\alpha\in (b_1,b_2): \limsup\limits_{x\rightarrow\infty}\pi^{\gamma*}_{\alpha,\beta}(x)\log^2x/x^{\gamma}<1-1/n \Big\}. $$ Obviously $\Cal{F}=\bigcup_{n>1}\Cal{F}_n$. So it suffices to show $\mes(\Cal{F}_n)=0$ for every $n>1$. Assume on the contrary that there exists $n>1$ such that $\mes(\Cal{F}_{n})>0$. Let $I:=(c_1,c_2)$ be an arbitrary subinterval of $(b_1,b_2)$, and let $P$ denote the set of Piatetski\hbox{-}Shapiro primes, we have $$ \aligned \int\limits_{c_{1}}^{c_{2}}\pi^{\gamma*}_{\alpha,\beta}(x)d\alpha &= \int\limits_{c_{1}}^{c_{2}} \pmatrix \sum\limits \Sb p\leq x\\p\in P\endSb \ \sum\limits \Sb \alpha p+\beta-11$ is the implied constant in Lemma 8. %\?Lemma 8 net, \Par*{Lemma 8} Let $\Cal{L}_I=\Cal{F}_{n}\cap I$ and $$ \Cal{L}_{I,\delta}(x)= \{\alpha\in I: \pi^{\gamma*}_{\alpha,\beta}(x)\log^2x/x^{\gamma}<1-\delta\}. $$ For any two primes $p$ and $q$, $$ J_{p.q}:=\{\alpha\in I:\lfloor\alpha p+\beta\rfloor=q\} $$ is an interval or an empty set. Let $$ \Cal{H}_{I,\delta}(x)= \pmatrix \bigcup\limits\Sb k>(1-\delta)x^{\gamma}/(\log x)^2 \\ p_1,\dots,p_k\leq x \text{ are distinct and } p_i\in P\\ q_1,\dots,q_k \text{ are primes} \endSb \bigcap\limits_{j=1}^{k}J_{p_j,q_j} \endpmatrix. $$ So we have that $$ \Cal{L}_{I,\delta}(x)=I\setminus\Cal{H}_{I,\delta}(x) $$ is measurable. By Lemma 8, %\?Lemma 8 net $$ \int\limits_{c_1}^{c_2}\pi^{\gamma*}_{\alpha,\beta}(x)d\alpha\leq \mes(\Cal{L}_{I,\delta}(x))\frac{(1-\delta)x^{\gamma}}{(\log x)^2}+(c_2-c_1-\mes(\Cal{L}_{I,\delta}(x)))\frac{Cx^{\gamma}}{(\log x)^2}. $$ Combining the above inequality with \Tag(12), we have $$ \mes(\Cal{L}_{I,\delta}(x))\leq\frac{C-1}{C-1+\delta}\mes(I). \tag12 $$ We claim that $$ \Cal{L}_{I}= \bigcap\limits_{m>n} \bigcup\Sb y\geq1\\y\in \bold{Z}\endSb \bigcap\Sb x\geq y\\x\in \bold{Z}\endSb \Cal{L}_{I,1/n-1/m}(x). \tag13 $$ In fact, for any $m>n$, if $$ \limsup\limits_{x\rightarrow\infty}\pi^{\gamma*}_{\alpha,\beta}(x)\log^2x/x^{\gamma}<1-1/n+1/m, $$ then there exists $y_0$ such that for any $x\geq y_0$ $$ \pi^{\gamma*}_{\alpha,\beta}(x)\leq(1-1/n+1/m)x^{\gamma}/(\log^2x). $$ On the other hand, if $\alpha\in \cup_y\cap_{x\geq y}\Cal{L}_{I,1/n-1/m}(x)$, clearly we have $$ \limsup\limits_{x\rightarrow\infty}\pi^{\gamma*}_{\alpha,\beta}(x)\log^2x/x^{\gamma}<1-1/n+1/m. $$ By \Tag(13) and (14), %\?нет \Tag(14), куда поставить или надо \Tag(12) and \Tag(13) we get $$ \mes(\Cal{L}_{I})\leq\limsup\limits_{x\rightarrow\infty}\mes(\Cal{L}_{1,1/3n}(x))\leq\frac{C-1}{C-1+1/3n}\mes(I). $$ Since $\mes(\Cal{F}_{n})>0$, there exist open intervals $I_1$, $I_2$, $\cdots$ $\subseteq(b_1,b_2)$ such that $$ \Cal{F}_{n}\subseteq\bigcup\limits_{k=1}^\infty I_k $$ and $$ \sum\limits_{k=1}^\infty \mes(I_k)\leq\frac{C-1+1/4n}{C-1}\mes(\Cal{F}_{n}). $$ But by \Tag(13), $$ \mes(\Cal{F}_{n})\leq\sum\limits_{k=1}^\infty \mes(\Cal{L}_{I_{k}})\leq\frac{C-1}{C-1+1/3n}\sum\limits_{k=1}^\infty \mes(I_k)\leq\frac{C-1+1/4n}{C-1+1/3n}\mes(\Cal{F}_{n}). $$ This leads to a contradiction. \acknowledgments We thank the anonymous referee for very helpful comments that improved the quality of this paper. %We thank the anonymous referee for his or her very useful comments on this paper. \Refs \ref\no 1 \by Li~H. and Pan~H. \paper Primes of the form $\lfloor\alpha p+\beta\rfloor$ \jour J. 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