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 Jan3 awarded Scholar Jan3 accepted Do regular partitions suffice for Riemann integrability of a real-valued function on a closed interval? Jan3 comment Do regular partitions suffice for Riemann integrability of a real-valued function on a closed interval? I just obtained full-text access to reference #2, where I could read the cited Theorem 3. And yes, the condition I asked about in my preceding comment is indeed the one shown there to suffice. Thanks! Jan3 comment Do regular partitions suffice for Riemann integrability of a real-valued function on a closed interval? I cannot access reference #2. For reference #1, I cannot access the relevant section beyond page 372. So would you mind quoting the theorem(s) to which you refer? And note that the Riemann-Darboux criterion in Theorem 2 of #2 involves controlling the difference between the Riemann sums using the maximum value and minimum value, respectively, of $f$ in each subinterval; but that's not the condition I asked about. Jan3 comment Do regular partitions suffice for Riemann integrability of a real-valued function on a closed interval? I'm not sure what the final sentence means. Does it mean the following? $f$ is integrable if: there exists some sequence of partitions $P_n$, with mesh tending to 0, for which, no matter what sample points in the subintervals of $P_n$ are used, the corresponding sequence of Riemann sums converges, and that limit is independent of the sample points chosen. (I knew the problem wasn't banal; I just didn't say what I meant the first time 'round.) Jan2 comment Do regular partitions suffice for Riemann integrability of a real-valued function on a closed interval? Taking the min $m$ of the non-regular partition's lengths is what I immediately thought of before I asked the question. The trouble is that you won't necessarily get a regular partition from it: $\Delta x = (b-a)/m$ need not be an integer. So the final subinterval need not have the same length as the others. Jan2 revised Do regular partitions suffice for Riemann integrability of a real-valued function on a closed interval? added 122 characters in body Jan2 comment Do regular partitions suffice for Riemann integrability of a real-valued function on a closed interval? Sorry, I meant to allow arbitrary choice of sample points in the subintervals, not just existence of sample points there. I've edited to indicate that. Yes, of course the characteristic function of the rationals on the unit interval is a counterexample to what I originally wrote; in fact, as I had in mind in my 2nd note, it's the well-known counterexample to allowing only endpoints as sample points. Jan2 awarded Editor Jan2 revised Do regular partitions suffice for Riemann integrability of a real-valued function on a closed interval? Changed existence of sample points to arbitrary choice of sample points. Jan2 asked Do regular partitions suffice for Riemann integrability of a real-valued function on a closed interval? Dec27 comment Easy example why complex numbers are cool @Mario Carneiro: My $(x - 2)$ was a typo, caught too late to edit. My point was not that determining the radius of convergence of some series expansions could not be facilitated by using complex numbers. Rather, my point was simply that the originally proffered example of $\frac{1}{1+x^2}$ was treatable easily by purely real methods. In any case, it's doubtful that power series expansion provides examples "explainable to someone only knowing high school mathematics" (which is what the OP requested). Dec27 comment Easy example why complex numbers are cool @Mario Carneiro: Expanding $1/(1+(x-2)^2)$ around what center? Dec26 comment Easy example why complex numbers are cool @MathNoob: The OP specifically asked for an example "explainable to someone only knowing high school mathematics." [emph added] Dec26 comment Easy example why complex numbers are cool That the radius of convergence of the real function $\frac{1}{1+x^2}$ is 1 hardly requires using poles of the corresponding complex function. Rewrite as $\frac{1}{1-(-x^2)}$ and note that $\lvert -x^2 \rvert = \lvert x\rvert^2$; but the radius of convergence of the geometric series expansion of $\frac{1}{1-x}$ is 1. Dec12 comment Genius mathematicians who never published anything @Michael Hardy: could you indicate the gist of Dame Cartwright's proof? (Although back in graduate school I heard her lecture about her work on dynamical systems growing out of the British radio problem, I never before heard about the proof you mention.) Dec7 comment What is a simple example of an unprovable statement? There's a second alternative: neither $X$ nor $Y$ is bigger than the other! Dec7 comment Why are integers subset of reals? 2 and 2.0 are the same mathematically because the latter means $2 + 0/10$. Dec1 comment An open set in $\mathbb{R}$ is a union of balls of rational radius and rational center. I think you're confusing that with the fact that $\mathbb{R}$ and, more generally, any separable metrizable space, is second-countable. Look at the proof of that. Dec1 comment “Honest” introductory real analysis book Exactly what real analysis topics do you expect to cover? Presumably either construction of the reals from integers or rationals, or at least an axiomatic treatment (which ought to include uniqueness up to order isomorphism); convergence of sequences and numerical series along with Taylor series and, more generally, convergence & uniform convergence of sequences and series of functions. The critical issue is which integral: just the Riemann integral? Riemann-Stieltjes? or Lebesgue? The 1st edition just did Riemann-Stieltjes integration, as I recall; the current, 3rd edition includes Lebesgue.