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let be $a:I\to K$, $J\subseteq I$, $J$ is finite, $K$ is Monoid. I define:

$\displaystyle \sum_J a:=\begin{cases} a(B) + \displaystyle \sum_{J-\{B\}} a & \text{, if } \exists x \in J:(x=B) \\ 0 & \text{, otherwise } \end{cases}$

Is it correct?

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    Why not $\sum_{i \in J} a_i $? – 2017-01-18
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    because I think $\sum_{i \in J} a_i$ is informal! – 2017-01-18
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    I think $∑_{i ∈ J} a_i$ is a well defined notation. The problem is that it doesn't address the definition part of your question. – 2017-01-21

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Maybe the induction in the definition would be more explicit if you first fixed an enumeration of $J$ and then formulated the induction using this enumeration.

Also note that it is not necessary to define the sum for subsets of $I$ since it is an instance of the definition for whole $I$.

You just want to formally define the sum of a finite function to a monoid. I would proceed as follows.

First, define $S(f)$ for every $f: n + 1 → K$ as $S(f) := S(f\restriction n) + f(n)$ and $0$ for $f: 0 → K$.

Then, define $∑a$ as $S(a \circ σ)$ for any bijection $σ: \lvert I\rvert → K$. (And prove that the choice of $σ$ doesn't matter.)

Finally, define $∑_J a$ as $∑(a\restriction J)$.

Also note that if $K$ was a topological monoid, you would be able to extend to the definition to infinite functions as well.