I know but it feels so weird
Yeah so let $S_0$ be $\{x\in A:f(x)\ge1\}$
Let $S_1$ be $\{x\in A:1>f(x)\ge\frac12\}$
Let $S_2$ be $\{x\in A:\frac12>f(x)\ge\frac13\}$
and in general let $S_n$ be $\{x\in A:\frac1n>f(x)\ge\frac1{n+1}\}$
We want to show that, if $m(A)>0$, and $f(x)>0$ for all $x$, then $\int_Af(x)dx>0$
The point is that $\bigcup\limits_{n=0}^\infty S_n=A$
So $\sum\limits_{n=0}^\infty m(S_n)=m(A)$
In particular, we can't have all the $m(S_n)$ be zero, because then their sum would be zero
So at least one of the $m(S_n)$ is greater than zero
So now: $\int_Af(x)dx\ge\int_{S_n}f(x)dx$
And, on $S_n$, $~f(x)$ is between $\frac1n$ and $\frac1{n+1}$ (so in particular it's greater than $\frac1{n+1}$)
so $\int_{S_n}f(x)dx\ge\int_{S_n}\frac1{n+1}dx=m(S_n)\frac1{n+1}>0$
The main point is that we can write the set of positive values as a union of countably many sets with positive lower bound
($(0,\infty)=[1,\infty)\cup[\frac12,1)\cup[\frac13,\frac12)\cup[\frac14,\frac13)\cup\dotsb$)
@Rithaniel Well, $(M,x)$ is always maximal
(It's the set of polynomials whose constant term is in $R$)
So the hard part would be the converse, yeah?
Right so if $\mathfrak M$ were not of the form $(M,x)$ then we need to show that there's some proper ideal strictly containing it, yeah?
Does it matter that it's $R[[x]]$ and not $R[x]$?
Thinking aloud for a second: What's the ideal in $\Bbb Q[x,y]$ generated by $x^n+y^n$, $~n=1,2,3,\dots$?
