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

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!

• What does the notation $E^n$ mean here, and what is a "closed interval of/in" it? Are you looking for a proof that every real-valued function on such an interval is integrable, or that there exists a real-valued function on some/every such interval that is integrable? – Henning Makholm Nov 19 '15 at 19:24
• Here $E$ refers to a metric space of $n$ dimensions. I was trying to prove this for all real valued function. And when I said closed interval, I was referring to a subset of $E^n$, such that the set contains its limit points ($[0,1]$ would be a closed interval of $E^1$). – Meecolm Nov 19 '15 at 19:29
• I'm not sure how you have defined integrability on general metric spaces, but in the good old real case, there are certainly functions on closed intervals that are not integrable, such as $$f:[0,1]\to\mathbb R \qquad f(x)=\begin{cases} 0 & \text{when }x=0 \\ 1/x & \text{otherwise}\end{cases}$$ Is this somehow excluded from your concept of a "real-valued function on a closed interval in $E^n$"? How? – Henning Makholm Nov 19 '15 at 19:33
• I cannot believe I forgot, the function has to be continuous! Otherwise, you are completely right. – Meecolm Nov 19 '15 at 19:36

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.