Sum of Neighborhoods of Zero When do two neighborhoods of zero over a topological vector space add up as:
$$aN+bN=(a+b)N\quad a,b\geq 0$$
I could imagine something like balanced might suffice...
The problem is that I'd like to estimate:
$$N|\mu(A_1)|+\ldots+N|\mu(A_N)|=N(|\mu(A_1)|+\ldots+|\mu(A_N)|)$$
 A: The equation $aN+bN=(a+b)N$ is true for any convex set $N$ in a real vector space when $a$ and $b$ have the same sign.
Clearly $aN+bN\supset(a+b)N$, so only the other direction needs to be shown.
Take $x\in aN+bN$.
Then there are $y,z\in N$ so that $x=ay+bz=(a+b)\left(\frac{a}{a+b}y+\frac{b}{a+b}z\right)$.
By convexity $\frac{a}{a+b}y+\frac{b}{a+b}z\in N$, so $x\in(a+b)N$.
Thus $aN+bN\subset(a+b)N$.
Remarks:


*

*The location of $N$ is irrelevant.
If you replace $N$ with $c+N$, the equation $aN+bN=(a+b)N$ is unchanged.

*If $N$ is balanced and convex, then the previous proof shows that $aN+bN=(|a|+|b|)N$.
If $a$ and $b$ have different sign and $N$ is not the whole space, then $aN+bN\neq(a+b)N$.
(Thus for bounded, balanced, convex sets (containing more than one point) OP's property holds if and only if $a$ and $b$ have the same sign.)

*Topology plays no role. This seems to be all about the vector space structure.


Edit:
The OP added to the question that they want to have the estimate
$$
N|\mu(A_1)|+\ldots+N|\mu(A_N)|=N(|\mu(A_1)|+\ldots+|\mu(A_N)|)\subseteq N|\mu|(A_1\cup\ldots\cup A_N),
$$
where $N$ is presumably balanced and convex.
Since the numbers $|\mu(A_i)|$ are all positive and thus have the same sign, the equality part is true.
The inclusion is equivalent with
$$
|\mu(A_1)|+\ldots+|\mu(A_N)|
\leq
|\mu|(A_1\cup\ldots\cup A_N).
$$
Establishing this requires more assumptions and it is false as stated (assuming $\mu$ is a measure, signed or not).
If $\mu$ is the Lebesgue measure on the real axis (now $\mu=|\mu|$) and $A_i=[i,3i]$, then $|\mu(A_1)|+|\mu(A_2)|=6>5\mu(A_1\cup A_2)$.
If the sets $A_i$ are disjoint and measurable (and $\mu$ is a signed measure), then
$$
|\mu(A_1)|+\ldots+|\mu(A_N)|
\leq
|\mu|(A_1)+\ldots+|\mu|(A_N)
=
|\mu|(A_1\cup\ldots\cup A_N),
$$
and the desired estimate is true.
