$\newcommand\de\delta\newcommand\ep\varepsilon\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$Saúl RM proved the desired result by an application of Frostman's lemma.

Here is an elementary proof:

Take any $\de$ and $\ep$ in $(0,1)$. Then, by the condition $H^{n-1}(K)=0$ and the definition of the Hausdorff measure, there is a set $\{Q_i\colon i\in\N\}$ of closed $n$-cubes in $\R^n$ such that $\bigcup_{i\in\N}Q_i\supseteq K$, $0<l(Q_i)<\de$ for all $i\in\N$, and \begin{equation*} \sum_{i\in\N}l(Q_i)^{n-1}<\ep/2, \tag{1}\label{1} \end{equation*} where $l(Q_i)$ is the edge length of the $n$-cube $Q_i$.

For each $i\in\N$, let $N_i:=\lceil 1/l(Q_i)\rceil$, so that \begin{equation*} \frac1{l(Q_i)}\le N_i\le1+\frac1{l(Q_i)}<\frac 2{l(Q_i)}. \tag{2}\label{2} \end{equation*} For each $i\in\N$ and each $j\in[N_i]:=\{1,\dots,N_i\}$, consider the $(n+1)$-cube \begin{equation*} R_{i,j}:=Q_i\times[(j-1)l(Q_i),jl(Q_i)], \end{equation*} with edge length $l(R_{i,j})=l(Q_i)<\de$. Then $\bigcup_{i\in\N}\bigcup_{j\in[N_i]}R_{i,j}\supseteq K\times[0,1]$ and \begin{equation*} \sum_{i\in\N}\sum_{j\in[N_i]}l(R_{i,j})^n =\sum_{i\in\N}\sum_{j\in[N_i]}l(Q_i)^n =\sum_{i\in\N}N_i l(Q_i)^n \\ <\sum_{i\in\N}\frac 2{l(Q_i)}l(Q_i)^n =2\sum_{i\in\N}l(Q_i)^{n-1}<\ep, \end{equation*} by \eqref{2} and \eqref{1}.

So, again by the definition of the Hausdorff measure, \begin{equation} H^n(K\times[0,1])=0. \end{equation} So/similarly, $H^n(K\times[k,k+1])=0$ for all integers $k$.

So, by the subadditivity of the Hausdorff measure, $H^n(K\times\R)=0$. $\quad\Box$

(The condition that $K$ is compact was not needed or used here.)

$\newcommand\de\delta\newcommand\ep\varepsilon\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$Saúl RM proved the desired result by an application of Frostman's lemma.

Here is an elementary proof:

Take any $\de$ and $\ep$ in $(0,1)$. Then, by the condition $H^{n-1}(K)=0$ and the definition of the Hausdorff measure, there is a set $\{Q_i\colon i\in\N\}$ of closed $n$-cubes in $\R^n$ (that are product sets) such that $\bigcup_{i\in\N}Q_i\supseteq K$, $0<l(Q_i)<\de$ for all $i\in\N$, and \begin{equation*} \sum_{i\in\N}l(Q_i)^{n-1}<\ep/2, \tag{1}\label{1} \end{equation*} where $l(Q_i)$ is the edge length of the $n$-cube $Q_i$.

For each $i\in\N$, let $N_i:=\lceil 1/l(Q_i)\rceil$, so that \begin{equation*} \frac1{l(Q_i)}\le N_i\le1+\frac1{l(Q_i)}<\frac 2{l(Q_i)}. \tag{2}\label{2} \end{equation*} For each $i\in\N$ and each $j\in[N_i]:=\{1,\dots,N_i\}$, consider the $(n+1)$-cube \begin{equation*} R_{i,j}:=Q_i\times[(j-1)l(Q_i),jl(Q_i)], \end{equation*} with edge length $l(R_{i,j})=l(Q_i)<\de$. Then $\bigcup_{i\in\N}\bigcup_{j\in[N_i]}R_{i,j}\supseteq K\times[0,1]$ and \begin{equation*} \sum_{i\in\N}\sum_{j\in[N_i]}l(R_{i,j})^n =\sum_{i\in\N}\sum_{j\in[N_i]}l(Q_i)^n =\sum_{i\in\N}N_i l(Q_i)^n \\ <\sum_{i\in\N}\frac 2{l(Q_i)}l(Q_i)^n =2\sum_{i\in\N}l(Q_i)^{n-1}<\ep, \end{equation*} by \eqref{2} and \eqref{1}.

So, again by the definition of the Hausdorff measure, \begin{equation} H^n(K\times[0,1])=0. \end{equation} So/similarly, $H^n(K\times[k,k+1])=0$ for all integers $k$.

So, by the subadditivity of the Hausdorff measure, $H^n(K\times\R)=0$. $\quad\Box$

(The condition that $K$ is compact was not needed or used here.)

$\newcommand\de\delta\newcommand\ep\varepsilon\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$Saúl RM proved the desired result by an application of Frostman's lemma.

Here is an elementary proof:

Indeed, take any $\de$ and $\ep$ in $(0,1)$. Then, by the condition $H^{n-1}(K)=0$ and the definition of the Hausdorff measure, there is a set $\{Q_i\colon i\in\N\}$ of closed $n$-cubes in $\R^n$ such that $\bigcup_{i\in\N}Q_i\supseteq K$, $0<l(Q_i)<\de$ for all $i\in\N$, and \begin{equation*} \sum_{i\in\N}l(Q_i)^{n-1}<\ep/2, \tag{1}\label{1} \end{equation*} where $l(Q_i)$ is the edge length of the $n$-cube $Q_i$.

For each $i\in\N$, let $N_i:=\lceil 1/l(Q_i)\rceil$, so that \begin{equation*} \frac1{l(Q_i)}\le N_i\le1+\frac1{l(Q_i)}<\frac 2{l(Q_i)}. \tag{2}\label{2} \end{equation*} For each $i\in\N$ and each $j\in[N_i]:=\{1,\dots,N_i\}$, consider the $(n+1)$-cube \begin{equation*} R_{i,j}:=Q_i\times[(j-1)l(Q_i),jl(Q_i)], \end{equation*} with edge length $l(R_{i,j})=l(Q_i)<\de$. Then $\bigcup_{i\in\N}\bigcup_{j\in[N_i]}R_{i,j}\supseteq K\times[0,1]$ and \begin{equation*} \sum_{i\in\N}\sum_{j\in[N_i]}l(R_{i,j})^n =\sum_{i\in\N}\sum_{j\in[N_i]}l(Q_i)^n =\sum_{i\in\N}N_i l(Q_i)^n \\ <\sum_{i\in\N}\frac 2{l(Q_i)}l(Q_i)^n =2\sum_{i\in\N}l(Q_i)^{n-1}<\ep, \end{equation*} by \eqref{2} and \eqref{1}.

So, again by the definition of the Hausdorff measure, \begin{equation} H^n(K\times[0,1])=0. \end{equation} So/similarly, $H^n(K\times[k,k+1])=0$ for all integers $k$.

So, by the subadditivity of the Hausdorff measure, $H^n(K\times\R)=0$. $\quad\Box$

(The condition that $K$ is compact was not needed or used here.)

$\newcommand\de\delta\newcommand\ep\varepsilon\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$Saúl RM proved the desired result by an application of Frostman's lemma.

Here is an elementary proof:

Take any $\de$ and $\ep$ in $(0,1)$. Then, by the condition $H^{n-1}(K)=0$ and the definition of the Hausdorff measure, there is a set $\{Q_i\colon i\in\N\}$ of closed $n$-cubes in $\R^n$ such that $\bigcup_{i\in\N}Q_i\supseteq K$, $0<l(Q_i)<\de$ for all $i\in\N$, and \begin{equation*} \sum_{i\in\N}l(Q_i)^{n-1}<\ep/2, \tag{1}\label{1} \end{equation*} where $l(Q_i)$ is the edge length of the $n$-cube $Q_i$.

For each $i\in\N$, let $N_i:=\lceil 1/l(Q_i)\rceil$, so that \begin{equation*} \frac1{l(Q_i)}\le N_i\le1+\frac1{l(Q_i)}<\frac 2{l(Q_i)}. \tag{2}\label{2} \end{equation*} For each $i\in\N$ and each $j\in[N_i]:=\{1,\dots,N_i\}$, consider the $(n+1)$-cube \begin{equation*} R_{i,j}:=Q_i\times[(j-1)l(Q_i),jl(Q_i)], \end{equation*} with edge length $l(R_{i,j})=l(Q_i)<\de$. Then $\bigcup_{i\in\N}\bigcup_{j\in[N_i]}R_{i,j}\supseteq K\times[0,1]$ and \begin{equation*} \sum_{i\in\N}\sum_{j\in[N_i]}l(R_{i,j})^n =\sum_{i\in\N}\sum_{j\in[N_i]}l(Q_i)^n =\sum_{i\in\N}N_i l(Q_i)^n \\ <\sum_{i\in\N}\frac 2{l(Q_i)}l(Q_i)^n =2\sum_{i\in\N}l(Q_i)^{n-1}<\ep, \end{equation*} by \eqref{2} and \eqref{1}.

So, again by the definition of the Hausdorff measure, \begin{equation} H^n(K\times[0,1])=0. \end{equation} So/similarly, $H^n(K\times[k,k+1])=0$ for all integers $k$.

So, by the subadditivity of the Hausdorff measure, $H^n(K\times\R)=0$. $\quad\Box$

(The condition that $K$ is compact was not needed or used here.)