(Substitution, part I) If $X$ is a finite set, $f: X \to R$ is a function, and $g: Y \to X$ is a bijection, then prove that $$\sum_{x \in X}f(x) = \sum_{y \in Y}f(g(y))$$
The definition may be helful:
The Definition: Let $X$ be a finite set with $n$ elements (where $n \in N$), and let $f: X → R$ be a function from $X$ to the real numbers (i.e., $f$ assigns a real number $f(x)$ to each element $x$ of $X$). Then we can define the finite sum $\sum_{x \in X}f(x)$ as follows. We first select any bijection $g$ from {$i ∈ N : 1 ≤ i ≤ n$} to $X$; such a bijection exists since $X$ is assumed to have $n$ elements. We then define $$\sum_{x \in X}f(x)=\sum_{i=1}^nf(g(i))$$
If $X$ is a finite set then there is a bijection from $h:${$i \in N: 1 \le i \le n$} $\to X$. Since $g$ is also a bijection to $X$, then we know that for some $j \in$ {$i \in N: 1 \le i \le n$}, $h(j) = g(y) $
$$\sum_{x \in X}f(x)=\sum_{i=1}^nh(i) = \sum_{j=1}^nh(j) = \sum_{y=1}^ng(y) = \sum_{y \in Y}f(g(y))$$
Where the very last equation holds by definition. Please, can you check the thoughts and if they are wrong help me to solve the problem?
$\sum_{y=1}^ng(y)$ doesn't make sense because $y \in Y$ and not $\{1,\dots,n\}$.
In order to define the sum $\sum_{y \in Y}f(g(y))$, you pick a bijection $h : \{1,\dots,n\} \to Y$. Then
$$ \sum_{y \in Y}f(g(y)) = \sum_{i = 1}^n f(g(h(i)). $$
Then the idea is that $g \circ h$ is a bijection $\{1,\dots,n\} \to X$ so
$$ \sum_{i = 1}^n f(g(h(i)) = \sum_{x \in X} f(x). $$