Consider sphere $B$ defined by the equation: $x^2+y^2+z^2 \le 1$. Is every continuous function defined on the sphere Riemann-integrable?
I have a hunch that the Jordan measure is the key to solving this problem. I know that if a function is continuous on a set and the set's boundary has measure $0$, then the function is integrable.
In this case, the boundary of the set is the set itself. And so, the boundary is the sphere $x^2 + y^2 + z^2 = 1$. Loosely speaking, we want the sphere to have $0$ volume. And I guess that is the case.
So, the answer to my question is affirmative?
In many textbooks the Riemann integral in ${\mathbb R}^n$ is first introduced for cubes $I:=[{-a},a]^n$ by means of Riemann sums. One then proves, among else, that for functions $f:\>I\to{\mathbb R}$ which are bounded on $I$ and continuous up to a set $N\subset I$ of $n$-dimensional Jordan measure $0$ the integral $\int_I f(x)\>{\rm d}(x)$ exists.
In order to integrate over arbitrary domains $B$ one defines $$\int_B f(x)\>{\rm d}(x):=\int_I 1_B(x)\> f(x)\>{\rm d}(x)\ .\tag{1}$$ If $f$ is continuous on $B$, and the boundary $\partial B$ has Jordan measure zero, then the integral $(1)$ exists.
In the case of a ball $B\subset{\mathbb R}^3$ the boundary $\partial B$ is a sphere. Such a sphere can be covered with finitely many patches which are $C^1$-images of compact rectangles in the $(u,v)$-plane. Such patches then have three-dimensional Jordan measure $0$.