Prove that a real-valued function in $n$ dimensions is integrable.`

Question

Prove that a continuous, real-valued function on a closed interval in $E^n$ is integrable.

Is it possible to deduct the above statement only using the following:

1. A real valued function function $f$ on a closed interval $I$ of $E^n$ is integrable on $I$ if and only if, given any $\epsilon > 0$, there exists a number $\delta > 0$ such that $|S_1 - S_2| < \epsilon$ whenever $S_1$ and $S_2$ are Riemann sums for $f$ corresponding to partitions of $I$ of width less than $\delta$.

2. A step function on a closed interval $I$ in $E^n$ is integrable. In particular, if $(x_1^0 , x_1^1, ... , x_1^{N_1}),...,(x_n^0 , x_n^1, ... , x_n^{N_n})$ is a partition of $I$, if $\{c_{j_1...j_n}\}_{j_1=1,...,N_1;...;j_n =1,...,N_n} \subset \Bbb R$, and if the step function $f:I \to \Bbb R$ is such that for any $j_1 = 1,...,N_1;...;j_n = 1,..., N_n$ we have $f(x_1, ..., x_n) = c_{j_1...j_n}$ if $x_i^{j_i-1} < x_i < x_i^{j_i}$ for each $i = 1,..., n$ then

$$ \int_I f = \sum_{j_1=1,...,N_1;...;j_n =1,...,N_n} c_{j_1...j_n}(x_1^{j_1} - x_1^{j_1-1}) \cdots(x_n^{j_n} - x_n^{j_n-1}) $$.

EDIT: You may also use uniform continuity!

Answer 1

We need to know what definition you are using to integrate functions on $E^n$ if we are to answer your question. For instance, if $E $ is the closed interval $[0,1] $ with the usual metric and the counting measure, then the constant function $f (x)=1$ is continuous, but has infinite integral.

If $E^n $ is $n $-dimensional Euclidean space, then products of closed intervals are compact, so any continuous function on such a product is bounded and thus has bounded Lebesgue integral.