Let a,b,cβR with aβ€bβ€c, and let f:[a,c]βR be Riemann integrable on [a,c]. We prove that the restrictions of f to [a,b] and [b,c] are Riemann integrable and that
β«acβf(t)dt=β«abβf(t)dt+β«bcβf(t)dt.
First we describe concatenation. Let
T=(a=x0β<β―<xmβ=b;Ο1β,β¦,Οmβ)
be a tagged partition of [a,b], and let
S=(b=y0β<β―<ynβ=c;Ο1β,β¦,Οnβ)
be a tagged partition of [b,c]. Their concatenation is the tagged partition of [a,c] obtained by joining the partition points and retaining the same tags on each subinterval. Its Riemann sum is exactly
R(f,TβS)=R(f,T)+R(f,S).
Let
I=β«acβf(t)dt.
We first show that fβ£[a,b]β is Riemann integrable on [a,b]. Fix Ξ΅>0. Since f is Riemann integrable on [a,c], there exists Ξ΄>0 such that every tagged partition of [a,c] of mesh less than Ξ΄ has Riemann sum within Ξ΅ of I.
Choose once and for all a tagged partition S0β of [b,c] with mesh less than Ξ΄, and write
J=R(f,S0β).
If T1β and T2β are tagged partitions of [a,b] with mesh less than Ξ΄, then the concatenations T1ββS0β and T2ββS0β are tagged partitions of [a,c] with mesh less than Ξ΄. Hence
β£R(f,T1ββS0β)βIβ£<Ξ΅,β£R(f,T2ββS0β)βIβ£<Ξ΅.
Using the concatenation identity, we obtain
β£R(f,T1β)βR(f,T2β)β£=β£R(f,T1ββS0β)βR(f,T2ββS0β)β£<2Ξ΅.
Thus the Riemann sums for fβ£[a,b]β over sufficiently fine tagged partitions form a Cauchy family in R, so they converge to some number I1ββR. It follows from the definition Riemann Integrability on a Closed Interval that fβ£[a,b]β is Riemann integrable on [a,b] and
I1β=β«abβf(t)dt.
The same argument, fixing a tagged partition of [a,b], shows that fβ£[b,c]β is Riemann integrable on [b,c]; write
I2β=β«bcβf(t)dt.
Finally, let Ξ·>0. Choose tagged partitions T of [a,b] and S of [b,c] of sufficiently small mesh such that
β£R(f,T)βI1ββ£<Ξ·,β£R(f,S)βI2ββ£<Ξ·,
and also the concatenation TβS has Riemann sum within Ξ· of I. Then
β£Iβ(I1β+I2β)β£β€β£IβR(f,TβS)β£+β£R(f,T)βI1ββ£+β£R(f,S)βI2ββ£<3Ξ·.
Since Ξ·>0 was arbitrary, one has I=I1β+I2β. Therefore
β«acβf(t)dt=β«abβf(t)dt+β«bcβf(t)dt.
Subtracting β«abβf(t)dt from both sides gives
β«acβf(t)dtββ«abβf(t)dt=β«bcβf(t)dt.