Show that $\sum_{j \in J} \sum_{i\in I_j}f(i) = \sum_{i\in I}f(i)\cdot \operatorname {card}\{j \in J: i \in I_j\}$

Let I be a finite indexing set and $f :I \rightarrow \mathbb{R}$ be a function.

We know that if $(I_j)_{j \in J}$ is a partition of I, then $\sum_{j \in J} \sum_{i\in I_j}f(i) = \sum_{i \in I}f(i)$

I'm wondering if it's true that if $I =\cup_{j \in J}I_j$ (not necessarily a disjoint union), then:

• $\sum_{j \in J} \sum_{i\in I_j}f(i) = \sum_{i\in I}f(i)\cdot \operatorname {card}\{j \in J: i \in I_j\}$

I'm almost convinced that this is true, but I don't know how to prove it, could anyone help me?

I'm trying to use this to prove this