If $S$ is a non-empty bounded set in $\mathbb{R}^N$, and $f,g:S\longrightarrow\mathbb{R}$ are Riemann integrable, show $fg$ is Riemann integrable.

I am not too sure with the $N$ dimensional proof of this, and have only worked it out in $1$-dimension.

My work so far is:

Suppose $f$ is bounded and Riemann integrable on $[a,b]$, then clearly $|f(x)|<a$ for some bound $a\geq 0$

So, $$|f^2(x)-f^2(y)|=|f(x)+f(y)||f(x)-f(y)|\leq2a|f(x)-f(y)|$$

By letting $M(f):=\sup\{f:x_{i-1}\leq x\leq x_i\}$ and $m(f):=\inf\{f:x_{i-1}\leq x \leq x_i\}$,

We see, $M(f^2)-m(f^2)\leq 2a(M(f)-m(f))$

By Cauchy Criterion, because $f$ is integrable on $[a,b]$, $\forall \epsilon > 0$, $\exists$ a partition $P$, such that

$$U(P,f)-L(P,f)< \frac{\epsilon}{2a}$$

$$\therefore U(P,f^2)-L(P,f^2)< \epsilon$$

Thus $f^2$ is Riemann integrable.

The claim then follows by noting $fg=\frac 12((f+g)^2-f^2-g^2)$

Is this the right way to proceed for a multivariable proof of the statement in n-dimensions? Or am I not approaching this correctly? How would my proof change for a multivariable version?

Any help would be appreciated. Thanks in advance!


That's just fine. There is indeed nothing different about this proof when you from one variable to several.


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