# Does integrating with respect to a finitely additive measure respect addition?

Let $X$ be a set and $\mathcal{A} \subseteq P(X)$ a $\sigma$-algebra. Assume $\nu : \mathcal{A} \to [0,\infty]$ is a finitely additive measure. If $f : X \to [0,\infty]$ is a measurable function, we can define $$\int_{X}f\,d\nu$$ in the standard way. If $f,g :X \to [0,\infty]$ are simple measurable functions then it is easy to prove that $$\int f + g\,d\nu = \int f\,d\nu + \int g\,d\nu.$$ However, if $f$ and $g$ are just measurable functions, then it is only obvious that $$\int f\,d\nu + \int g\,d\nu \leq \int f + g\, d\nu.$$

Question : Does integration with respect to a finitely additive measure respect addition?

Note, that if $\nu$ is countably additive, then the standard way to prove that integration respects addition is to appeal to the monotone convergence theorem.

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Yes, you can show that it is additive reasonably directly from the definitions. But, I think this must be a standard result for finitely additive measures, right? –  George Lowther Nov 1 '11 at 1:39
@George: If you can show it directly from the definitions then I am missing something. I have been thinking about this problem for a few hours and not made much progress. Also thanks for correcting the typos, I forgot that I was using a machine that does not have spell check enabled ;) –  Daniel Barter Nov 1 '11 at 1:53

As mentioned in the question, the inequality \begin{align}\int(f+g)\\,d\nu\ge\int f\\,d\nu+\int g\\,d\nu&&{\rm(1)}\end{align} follows easily from the definition of the integral $\int f\\,d\nu$ (as the supremum of the integrals of nonnegative simple functions bounded by $f$). So, I'll just show the reverse inequality which will establish additivity of the integral.
Choose any nonnegative simple function $h\le f+g$. We need to show that $\int f\\,d\nu+\int g\\,d\nu\ge\int h\\,d\nu$. However, we can write $h=\sum_{k=1}^nc_k1_{A_k}$ for $c_k\in(0,\infty)$ and pairwise disjoint $A_k\in\mathcal{A}$. As long as it can be shown that $\int_{A_k}f\\,d\nu+\int_{A_k}g\\,d\nu\ge c_k\nu(A_k)$, then the required inequality will follow by summing over $k$ and applying (1). So, replacing $f,g$ by $1_{A_k}f,1_{A_k}g$ respectively (for a fixed $k$), we reduce to the case where $n=1$. Dividing through by $c_k$ reduces to $c_k=1$.
So, we have reduced to the situation with $f+g\ge1_A$ and just need to show that $\int f\\,d\nu+\int g\\,d\nu\ge\nu(A)$. Without loss of generality (capping $f,g$ by 1 if necessary), we further reduce to the case with $0\le f,g\le1$. Then, for each positive integer $N$, consider the simple functions \begin{align} f_N&=\sum_{j=0}^{\lfloor N\rfloor}1_{f^{-1}((j/N,(j+1)/N])}\frac jN\le f,\\\\ g_N&=\sum_{j=0}^{\lfloor N\rfloor}1_{g^{-1}((j/N,(j+1)/N])}\frac jN\le g. \end{align} We have $f_N+g_N\ge(1-\frac2N)1_A$. So, using additivity for simple functions $$\int f\\,d\nu+\int g\\,d\nu\ge\int f_N\\,d\nu+\int g_N\\,d\nu\ge\left(1-\frac2N\right)\nu(A).$$ Letting $N$ increase to infinity gives the required inequality.