It is known that if $R$ is a Noetherian ring, then an element $x\in K(R)$ in its total quotient ring belongs to $R$ iff the image of $x$ in $K(R)_{\mathfrak{p}}$ belongs to $R_{\mathfrak{p}}$ for every prime ${\mathfrak{p}}$ associated to a nonzerodivisor in $R$. (For example, refer to Eisenbud; proposition 11.3.)
What I am curious of is whether the below sentence is true:
($R$ is Noetherian.) $R_{\mathfrak{p}}$ is integrally closed in its total quotient ring $K(R_{\mathfrak{p}})$ for every prime associated to a nonzerodivisor, then $R$ is integrally closed in $K(R)$.
I could not found such is known already so tried to prove under assumption it is correct. Here is the outline of my proof:
Let $x\in K(R)$. To show that $x\in R$. It is enough to show the image of $x$ in $K(R)_{\mathfrak{p}}$, say $x^*$, also belongs to $R_{\mathfrak{p}}$. But we did not assume that $R_{\mathfrak{p}}$ is integrally closed in $K(R)_{\mathfrak{p}}$ but $R_{\mathfrak{p}}$ is integrally closed in its total quotient ring. So it is necessary to show that a proper element in $K(R_{\mathfrak{p}})$ belongs to $R_{\mathfrak{p}}$ and it in real shows that $x^*$ belongs to $R_{\mathfrak{p}}$. Set an integral equation over $R_{\mathfrak{p}}$ for $x^*$ and induce an equation in $R$ just with the equivalence relations which is needed to construct localizations. And with that equation, I could again induce an integral equation over $R_{\mathfrak{p}}$ for some $y\in K(R_{\mathfrak{p}})$. INDEED, if $x^*$ can be written as $\frac{\frac{a}{u}}{1}$, where $\frac{a}{u}\in K(R)$, $y$ can be written as $\frac{\frac{a}{1}}{\frac{u}{1}}$. By using the fact $y$ belongs to $R_{\mathfrak{p}}$, I could show $x^*$ belongs to $R_{\mathfrak{p}}$.
IN summary I want to know:
Whether such sentence is well-known already...
True or not
If it is true, is there any other simpler proof than I have shown.
Edit: Assuming $R_{\mathfrak{p}}$ is domain for all such primes, this may be true. (But I am not sure.)