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How do we know that upper Riemann sum is greater than the area under a curve? Isn't it a bit circular?

What I mean is that we define area under a curve as a limit of upper and lower Riemann sums (assuming the respective limits are equal). But what motivates the definition of area under the curve is that we believe upper sums are always greater or equal to true area (less or equal in case of lower sums) - but we don't know what area is at this point (we define it later using Riemann sums), this is the circularity I mentioned.

Could anyone clarify this for me? It has to follow probably from some axioms of area. The first thing to do would be probably to show that the set of points under the graph is a subset of the set of points 'below' the upper Riemann sum.

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  • $\begingroup$ "But what motivates the definition of area under the curve is that we believe upper sums are always greater or equal to true area": isn't that a circular question? :) $\endgroup$ – jdoicj Oct 22 '15 at 11:24
  • $\begingroup$ We define what area is for rectangles first. This is the key. After that, since we want area to be monotone, we define area under the curve as the upper Riemann sum. $\endgroup$ – Crostul Oct 22 '15 at 11:27
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I would say this comes from the fact that we would like the area function to be "monotonic". That is, if $A \subset B$, we want $area(A) \leq area(B)$ (think about monotonicity of a measure in a measure space). So, given a positive bounded function on an interval $[a,b]$, it is a fact that the set of points under the curve is a subset of the union of the collection of rectangles determined by an upper sum. Hence, it's convenient to regard the area of curve to be less than or equal area of rectangles in an upper sum.

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  • $\begingroup$ First we need to prove that the set of points under bounded function on $[a,b]$ is a subset of the union of the collections of rectangles determined by an upper sum. How would you do that? $\endgroup$ – user4205580 Oct 22 '15 at 19:05
  • $\begingroup$ Let $P=\{a=p_{0}, p_1,\dots, p_n = b\}$ be the partition of $[a,b]$ we are considering. Let $M_k=sup\{ f(x): x \in [p_k,p_{k+1}]\}$. The rectangle corresponding to $[p_k,p_{k+1}]$ is the set $R_k=\{(x,y): x \in [p_k,p_{k+1}] \text{ and } y \in [0,M_k]\}$. Let $(x,f(x))$ for $x \in [a,b]$ and say $x \in [p_k,p_{k+1}]$. By definition of $M_k$, $f(x) \leq M_k$. Hence, $(x,f(x)) \in R_k$. $\endgroup$ – Roberto Nunez Oct 22 '15 at 23:55

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