$\newcommand{\N}{\mathbb{N}}\newcommand{\R}{\mathbb{R}}$There are examples on $\R$ with the Borel $\sigma$-algebra $\mathcal{B}$. We take the null ideal to be the meagre Borel sets $\mathcal{M}$ (the $\sigma$-ideal in the Borel sets generated by closed sets with empty interior).

The regular open sets of $\R$ form a complete Boolean algebra $\mathcal{RO}$, and the mapping from $\mathcal{RO} \rightarrow \mathcal{B}/\mathcal{M}$ formed by mapping a regular open set to the equivalence class of Borel sets differing from it by a meagre set is an isomorphism (this uses the Baire category theorem - see for example Fremlin's Measure Theory 514I). What we shall do is define a finitely additive measure $\mu$ on $\mathcal{RO}$ for which the only null element is $\emptyset$. Under the isomorphism above, this defines a finitely-additive Borel probability measure on $\R$ whose null ideal is $\mathcal{M}$.

Let $(U_i)_{i \in \N}$ be a countable base of regular open sets for $\R$ (e.g. open intervals with rational endpoints). By the ultrafilter lemma, for each $i \in \N$, there exists an ultrafilter on $\mathcal{RO}$ containing $U_i$, which defines a finitely-additive measure $\mu_i : \mathcal{RO} \rightarrow [0,1]$ taking only the values $0$ and $1$ and such that $\mu_i(U_i) = 1$.

We then define $\mu : \mathcal{RO} \rightarrow [0,1]$ by $\mu(U) = \sum_{i=1}^\infty 2^{-i} \mu_i(U)$. It is easy to verify that this is a finitely-additive probability measure. Also, for any non-empty regular open $U$ there exists some $i \in \N$ such that $U_i \subseteq U$, and therefore $$ \mu(U) \geq \mu(U_i) \geq 2^{-i}\mu_i(U_i) = 2^{-i} > 0. $$ So the only $\mu$-null regular open set is $\emptyset$.

The measure $\mu$ is not countably additive because on Polish spaces there are no countably-additive Borel probability measures that vanish on meagre sets.

$\newcommand{\N}{\mathbb{N}}\newcommand{\R}{\mathbb{R}}$There are examples on $\R$ with the Borel $\sigma$-algebra $\mathcal{B}$. We take the null ideal to be the meagre Borel sets $\mathcal{M}$ (the $\sigma$-ideal in the Borel sets generated by closed sets with empty interior).

The regular open sets of $\R$ form a complete Boolean algebra $\mathcal{RO}$, and the mapping from $\mathcal{RO} \rightarrow \mathcal{B}/\mathcal{M}$ formed by mapping a regular open set to the equivalence class of Borel sets differing from it by a meagre set is an isomorphism (this uses the Baire category theorem - see for example Fremlin's Measure Theory 514I). What we shall do is define a finitely additive measure $\mu$ on $\mathcal{RO}$ for which the only null element is $\emptyset$. Under the isomorphism above, this defines a finitely-additive Borel probability measure on $\R$ whose null ideal is $\mathcal{M}$.

Let $(U_i)_{i \in \N}$ be a countable base of regular open sets for $\R$ (e.g. open intervals with rational endpoints). By the ultrafilter lemma, for each $i \in \N$, there exists an ultrafilter on $\mathcal{RO}$ containing $U_i$, which defines a finitely-additive measure $\mu_i : \mathcal{RO} \rightarrow [0,1]$ taking only the values $0$ and $1$ and such that $\mu_i(U_i) = 1$.

We then define $\mu : \mathcal{RO} \rightarrow [0,1]$ by $\mu(U) = \sum_{i=1}^\infty 2^{-i} \mu_i(U)$. It is easy to verify that this is a finitely-additive probability measure. Also, for any non-empty regular open $U$ there exists some $i \in \N$ such that $U_i \subseteq U$, and therefore $$ \mu(U) \geq \mu(U_i) \geq 2^{-i}\mu_i(U_i) = 2^{-i} > 0. $$ So the only $\mu$-null regular open set is $\emptyset$.

The measure $\mu$ is not countably additive because on Polish spaces without isolated points there are no countably-additive Borel probability measures that vanish on meagre sets.

$\newcommand{\N}{\mathbb{N}}\newcommand{\R}{\mathbb{R}}$There are examples on $\R$ with the Borel $\sigma$-algebra $\mathcal{B}$. We take the null ideal to be the meagre Borel sets $\mathcal{M}$ (the $\sigma$-ideal in the Borel sets generated by closed sets with empty interior).

The regular open sets of $\R$ form a complete Boolean algebra $\mathcal{RO}$, and the mapping from $\mathcal{RO} \rightarrow \mathcal{B}/\mathcal{M}$ formed by mapping a regular open set to the equivalence class of Borel sets differing from it by a meagre set is an isomorphism (this uses the Baire category theorem - see for example Fremlin's Measure Theory 514I). What we shall do is define a finitely additive measure $\mu$ on $\mathcal{RO}$ for which the only null element is $\emptyset$. Under the isomorphism above, this defines a finitely-additive Borel probability measure on $\R$ whose null ideal is $\mathcal{M}$.

Let $(U_i)_{i \in \N}$ be a countable base of regular open sets for $\R$ (e.g. open intervals with rational endpoints). By the ultrafilter lemma, for each $i \in \N$, there exists an ultrafilter on $\mathcal{RO}$ containing $U_i$, which defines a finitely-additive measure $\mu_i : \mathcal{RO} \rightarrow [0,1]$ taking only the values $0$ and $1$ and such that $\mu_i(U_i) = 1$.

We then define $\mu : \mathcal{RO} \rightarrow [0,1]$ by $\mu(U) = \sum_{i=1}^\infty 2^{-i} \mu_i(U)$. It is easy to verify that this is a finitely-additive probability measure. Also, for any non-empty regular open $U$ there exists some $i \in \N$ such that $U_i \subseteq U$, and therefore $$ \mu(U) \geq \mu(U_i) \geq 2^{-i}\mu_i(U_i) = 2^{-i} > 0. $$ So the only $\mu$-null regular open set is $\emptyset$.

The measure $\mu$ is not countably additive because there are no countably-additive Borel probability measures vanishing on meagre sets.

$\newcommand{\N}{\mathbb{N}}\newcommand{\R}{\mathbb{R}}$There are examples on $\R$ with the Borel $\sigma$-algebra $\mathcal{B}$. We take the null ideal to be the meagre Borel sets $\mathcal{M}$ (the $\sigma$-ideal in the Borel sets generated by closed sets with empty interior).

The regular open sets of $\R$ form a complete Boolean algebra $\mathcal{RO}$, and the mapping from $\mathcal{RO} \rightarrow \mathcal{B}/\mathcal{M}$ formed by mapping a regular open set to the equivalence class of Borel sets differing from it by a meagre set is an isomorphism (this uses the Baire category theorem - see for example Fremlin's Measure Theory 514I). What we shall do is define a finitely additive measure $\mu$ on $\mathcal{RO}$ for which the only null element is $\emptyset$. Under the isomorphism above, this defines a finitely-additive Borel probability measure on $\R$ whose null ideal is $\mathcal{M}$.

Let $(U_i)_{i \in \N}$ be a countable base of regular open sets for $\R$ (e.g. open intervals with rational endpoints). By the ultrafilter lemma, for each $i \in \N$, there exists an ultrafilter on $\mathcal{RO}$ containing $U_i$, which defines a finitely-additive measure $\mu_i : \mathcal{RO} \rightarrow [0,1]$ taking only the values $0$ and $1$ and such that $\mu_i(U_i) = 1$.

We then define $\mu : \mathcal{RO} \rightarrow [0,1]$ by $\mu(U) = \sum_{i=1}^\infty 2^{-i} \mu_i(U)$. It is easy to verify that this is a finitely-additive probability measure. Also, for any non-empty regular open $U$ there exists some $i \in \N$ such that $U_i \subseteq U$, and therefore $$ \mu(U) \geq \mu(U_i) \geq 2^{-i}\mu_i(U_i) = 2^{-i} > 0. $$ So the only $\mu$-null regular open set is $\emptyset$.

The measure $\mu$ is not countably additive because on Polish spaces there are no countably-additive Borel probability measures that vanish on meagre sets.