The following is a proof I thought of for the infinitude of prime numbers which is based on measure theory. I believe it is original. Do you spot a false step in the arguments below or is it a correct proof?
Proof for infinitude of primes:
We begin by defining a measure on over the Borel $\sigma$-algebra over the reals, $\mathscr{B}(\mathbb{R})$.
For a given integer n with prime decomposition $n=\Pi_{k=1}^{m(n)}p_{k}^{e_{k}(n)}\in\mathbb{N}$ , where $\{p\_k\}\_{k}$ is a count of primes, m(n) some large enough index and $e_{k}(n)$ are the decomposition exponents, we define the family of function $s_{\ell}:\mathbb{N}\rightarrow(0,1]$ by $$s_{\ell}(n)=s_{\ell}(\Pi_{k=1}^{m(n)}p_{k}^{e_{k}(n)})=\ell^{1-\sum_{k=1}^{m(n)}e_{k}(n)}$$ where $\ell$ is a positive real number. Therefore n is prime iff $s_{\ell}(n)=1$, otherwise $0<s_{\ell}(n)<1$.
Now define the family of functions $\mu_{\ell}:\mathscr{B}(\mathbb{R})\rightarrow[0,\infty]$ by$$\mu_{\ell}(I)=\sum_{n\in I\cap\mathbb{N}}s_{\ell}(n)$$
Now we show $\mu_{\ell}$ is a measure. First it is non-negative because $s_{\ell}$ is a positive function. Second, for the empty set $$\mu_{\ell}(\emptyset)=\sum_{n\in\emptyset\cap\mathbb{N}}s_{\ell}(n)=\sum_{n\in\emptyset}s_{\ell}(n)=0$$ And Last, let $\lbrace E_{i}\rbrace_{i=1}$ pairwise disjoint sets in $\mathscr{B}(\mathbb{R})$ then $$\mu_{\ell}(\cup_{i=1}^{\infty}E_{i})=\sum_{n\in\left(\cup_{i=1}^{\infty}E_{i}\right)\cap\mathbb{N}}s_{\ell}(n)=\sum_{n\in\cup_{i=1}^{\infty}\left(E_{i}\cap\mathbb{N}\right)}s_{\ell}(n)=\sum_{i=1}^{\infty}\sum_{n\in\left(E_{i}\cap\mathbb{N}\right)}s_{\ell}(n)=\sum_{i=1}^{\infty}\mu_{\ell}(E_{i})$$ as required.
Now we show the infinitude of primes.
Proof. Assume to the contrary that the number of primes is finite and denote its set $\{p_{i}\}_{i=1}^{N}$ with N the (finite) number of primes.
Define the following sets $$P^{k}=\lbrace n\in\mathbb{N}\|\sum_{k=1}^{N}e_{k}(n)=k\rbrace=\lbrace n\in\mathbb{N}\ |s_{\ell}(n)=\ell^{1-k}\rbrace$$ i.e. a given $P^{k}$ contains all natural numbers whose sum of exponents appearing in its (assumed) finite prime decomposition $n=\Pi_{k=1}^{N}p_{k}^{e_{k}(n)}$ equals to k. Note that for example $P^{1}$ is the set of all prime numbers, which is $\lbrace p_{i}\rbrace_{i=1}^{N}$ under our assumptions. $P^{2}$ is the set of integer constructed from multiplying two primes from $\lbrace p_{i}\rbrace_{i=1}^{N}$ and so on.
Some facts about the $P^{k}$'s. First, for a given k, one can construct an integer in $P^{k}$ by choosing any k primes from $\lbrace p_{i}\rbrace_{i=1}^{N}$ with repetition, therefore $|P^{k}|=N^{k}$. Second, if $n_{1}\in P^{k_{1}}$ and $n_{2}\in P^{k_{2}}$ it must be that $n_{1}\neq n_{2}$ since they must have a different prime decomposition which is unique, and so,$$P^{k}\text{ are pairwise disjoint}$$Last, since every prime decomposition of a number n has some sum $\sum_{k=1}^{N}e_{k}(n)$ we get that $$\mathbb{N}=\cup_{k=1}^{\infty}P^{k}$$ Now lets computer the measure of the set $\mathbb{N}\in\mathscr{B}(\mathbb{R})$ selecting the measure $\mu_{N+1}$:$$\mu_{N+1}(\mathbb{N}) =\mu_{N+1}(\cup_{k=1}^{\infty}P^{k})=\sum_{k=1}^{\infty}\mu_{N+1}(P^{k})=\sum_{k=1}^{\infty}\sum_{n\in P^{k}\cap\mathbb{N}}s_{N+1}(n) =\sum_{k=1}^{\infty}\sum_{n\in P^{k}}s_{N+1}(n)=\sum_{k=1}^{\infty}|P^{k}|(N+1)^{-k}=\sum_{k=1}^{\infty}\left(\frac{N}{N+1}\right)^{k} =\sum_{k=0}^{\infty}\left(\frac{N}{N+1}\right)^{k}-1=\frac{1}{1-\frac{N}{N+1}}-1=N<\infty$$ On the other hand, for every integer n with prime composition $n=\Pi_{k=1}^{N}p_{k}^{e_{k}}$ we have the following bound $$\log_{2}n=\sum_{k=1}^{N}e_{k}\log_{2}p_{k}\geq\log_{2}p_{1}\sum_{k=1}^{N}e_{k}=(\log_{2}2)\sum_{k=1}^{N}e_{k}=\sum_{k=1}^{N}e_{k}$$ And now recompute the measure,$$\mu_{N+1}(\mathbb{N}) =\sum_{n\in\mathbb{N}\cap\mathbb{N}}s_{N+1}(n) =\sum_{n=1}^{\infty}(N+1)^{1-\sum_{k=1}^{N}e_{k}(n)}\geq\sum_{n=1}^{\infty}(N+1)^{1-\log_{2}n} \geq2\sum_{n=1}^{\infty}2^{-\log_{2}n}=2\sum_{n=1}^{\infty}\frac{1}{n}=\infty$$ We encountered a contradiction, therefore the assumption is false and the theorem is true $\square$