Doubt in Integration a function on a $k$ - cell.

Suppose $I^k$ is a $k$-cell in $R^k$, consisting of all $X = (x_1, ... , x_k)$ such that $a_i \le x_i \le b_i (i = 1, \cdots , k)$,

$I^j$ is the $j$-cell in $R^j$ defined by the first $j$ inequalities, and $f$ is a real continuous function on $I^k$. (What does real continuous function in this case mean?- A function $f:I^k \to \mathbb{R}$?)

Put $f = f_k$, and define $f_{k-1}$ on $I^{k-1}$ by

$$f_{k-1}(x_1, \cdots , x_{k-1}) = \int_{a_k}^{b_k}h(x_1,\cdots, x_{k-1}, x_k) d_{x_k}\cdots (**)$$.

let us introduce the temporary notation $L(f)$ for the integral (**) and $L'(f)$ for the result obtained by carrying out the k integrations in some other order.

For every $f \in \mathbb{C}(I^k), L(f) = L'(f)$

Proof: If $h(x) = h_1(x_1)\cdots h_k(x_k)$,(I really don't understand how and why are we considering such a function h(x)) where $h_j \in \mathbb{C}([a_i, b_i])$,

$L(h) = \prod_{i = 1}^k \int_{a_i}^{b_i} h_i(x_i) d_{x_i} = L'(h)$. (How are we getting this?) Put $V = \prod_{i=1}^k(b_i-a_i)$

If $f \in \mathbb{C}(I^k)$ and $\epsilon > 0$, there exists $g \in \mathbb{A} $ such that $||f - g|| < \epsilon/ V$, where $||f||$ is defined as max $|f(x)|$ ($x \in I^k$). Then $|L(f-g)| < \epsilon$, $|L'(f-g)|< \epsilon$,(** I don't understand this calculation also**) and since

$L(f) - L'(f) = L(f - g) + L'(g - f)$, we conclude that $L(f) - L'(f) < 2\epsilon$.

Solution 1:

1. Yes, a real continuous function on a topological space $X$ is a continuous function $X\to\Bbb R$.

2. We consider the set $\Bbb A$ of functions of this specific form $h(x)=\prod_i h_i(x_i)$ for two reasons: first, we can calculate their integrals easily and they readily satisfy the statement (see point 3) and second, these functions are dense among all real functions on $I^k$. This second one is not a trivial statement, presumably proved beforehand, utilizing that $I^k$ is compact.

3. For clarity, let's verify it for $k=2$. The same (or induction) applies for arbitrary $k$. $$\int_{I^k}h_1(x_1)h_2(x_2)\,d(x_1,x_2)=\int_{a_1}^{b_1}\!\!\int_{a_2}^{b_2} h_1(x_1)h_2(x_2)\,dx_2dx_1= \int_{a_1}^{b_1}\left(h_1(x_1)\cdot \int_{a_2}^{b_2} h_2(x_2)\,dx_2\right)\,dx_1=\left(\int_{a_2}^{b_2} h_2(x_2)\,dx_2\right)\cdot\int_{a_1}^{b_1}h_1(x_1)\,dx_1$$

4. It's using the claim that our special functions in 2. are in fact dense among all continuous real functions, which means exactly that for any $f\in C(I^k)$ and $\varepsilon>0$ there's a function $g\in\Bbb A$ (of the special form) such that $\|f-g\|<\varepsilon$.