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In my calculus textboox appears the following statement about the Riemann integral:

Let R([a,b]) be the set of all Riemann integrable functions of [a,b].

i) If [c,d] ⊂ [a,b] and f ∈ R([a,b]), then f ∈ R([c,d]).

I am trying to see why f ∈ R([c,d]), but I can't seem to understand why should f belong to a subinterval of [a,b].

Albelaski
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    $f$ doesn't belong to a subinterval. You're just looking at the part of the graph of $f$ over the interval $[c,d]$. So if you have area under the curve on $[a,b]$, why do you have area under the curve under the subinterval $[c,d]$? [If you're trying to prove it, consider partitions of $[a,b]$ that use points $c$ and $d$.] – Ted Shifrin Oct 05 '17 at 18:35
  • See my answer to this question . – Tony Piccolo Oct 06 '17 at 16:05

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