If $X$ is dicrete and $m$ satisfies probability axioms except $m(\bigcup_{n\in J}A_n)=\sum_{n\in J}m(A_n)$ for finite $J$, does countable sum hold?

Question

Let $X\subset\mathbb{R}^n$ be a non-empty discrete set in the standard topology and $m:\mathcal{P}(X)\to[0,1]$ be a function satisfying all probability measure axioms except that countable summability for disjoint sets holds a priori only as a finite version, that $m\left(\bigcup_{n\in J}A_n\right) = \sum_{n\in J}m(A_n)$ for all finite index sets $J\subset\mathbb{N}$. Does $m$ then also satisfy the countable summability?

I suspect that if $|X| < \infty$, then the answer is yes for any countable union of non-empty disjoint sets of $X$ must cover $X$ after some index $K$. But other than that, I don't know. I have an ithcing that the answer is false, since proofs of this sort usually use at least the countable subadditivity of outer measures. But I am yet to produce or find a suitable counterexample.

What do you mean by "probability measure" axioms? . Do you mean the measure axioms? . Well if yes, then they are only three in number . Please define the properties you want $m$ to have. Many "properties" of probability measures are a consequence of countable additivity or sometimes they themselves imply countable additivity. So please edit your question. — Mr.Gandalf Sauron, Jan 11 '23 at 11:47
@Mr.GandalfSauron The post you linked answers my question perfectly. Do you want to post it as a proper question so that I can close my question? — Epsilon Away, Jan 11 '23 at 16:54

If $X$ is dicrete and $m$ satisfies probability axioms except $m(\bigcup_{n\in J}A_n)=\sum_{n\in J}m(A_n)$ for finite $J$, does countable sum hold?

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