ΘρϵηΠατπ

Partition
A partition of [a,b] is a finite tuple sorted in ascending order P=(x1,x2,xn+1) such that a=x1 and b=xn+1
Refinement of a Partition
A partition R is a refinement of a partition P if PR
Common Refinement
Suppose that P,Q are partitions, then R is a common refinement of P and Q if R is a refinement of PQ
i-th Section of a Partition
Suppose that P=(x1,xn+1) is a partition, then for each i[1,,n] we define si:=[xi,xi+1]
Supremum of a Bounded Function on a Section
Suppose that f:[a,b] and that P=(x1,,xn+1) is a partition of [a,b] then for each i[1,,n] we have Mi(f,P)=sup({f(x):x[xi,xi+1]})
i-th Delta of a Partition
Suppose that P=(x1,,xn+1) is a partition, then for each i[1,n] we define: Δi=xi+1xi

Note that if a partition has k elements then there will be k1 deltas.

A Delta Sum Telescopes
Suppose that P=(x1,,xn+1) is a partition of [a,b] then i=1nΔi=ba
Recall that i=1nΔi=i=1n(xi+1xi)=1xn+1x1=ba where we've used fact 1.
Mesh of a Partition
Suppose that P is a partition of [a,b] such that |P|=n+1 for some n0 then mesh(P)=max({Δi:i[1,,n]})
Upper Sum of a Bounded Function over a Partition
Suppose that f:[a,b] is bounded, and that P is a partition of [a,b] then we define U(f,P):=j=1nMj(f,P)Δj
Lower Sum of a Bounded Function over a Partition
Suppose that f:[a,b] is bounded, and that P is a partition of [a,b] then we define L(f,P):=j=1nmj(f,P)Δj
Upper Sum Decreases over Refinements
Suppose that PR then U(f,P)U(f,R)
Lower Sum Decreases over Refinements
Suppose that PR then L(f,P)L(f,R)
Riemann Integrable
Suppose that f:[a,b] is bounded, then we say that it is Riemann integrable if supP(L(f,P))=infQ(U(f,Q)) In that case we write abf(x)dx as their common value
Riemann Integrable Characterizations
Let f(x) be bounded on [a,b] then the following are equivalent
  • f is Riemann integrable
  • For each ϵ+ there is a partition P such that
  • U(f,P)L(f,P)<ϵ
  • For every ϵ+ there is some δ+ such that for every partition Q such that mesh(Q)<δ wherein
  • U(f,Q)L(f,Q)<ϵ
  • There exists an I such that for every ϵ+ there is a δ+ such that for every partition Q with mesh(Q)<δ and every evaluation sequence X of Q we have
  • |I(f,Q,X)I|<ϵ and in that case I=abf(x)dx

Note that bullet point 2 is a the most tractable for explicit functions, as one can construct a partition that works for a given function.

Jump Discontinuity
Suppose f:[a,b], and c(a,b) then if limxc+f(x)limxcf(x) then we say that f has a jump discontinuity at c
Function Difference Set is the Same as its Negative
Suppose that f:A and define F={f(x)f(y):x,yA}, then F=F
Note that F={f(x)f(y):x,yA}={(f(y)f(x)):y,xA}=F as needed.
Supremum of the Function Difference Set is the Same with Absolute Values
Let f:A and define F={f(x)f(y):x,yA} then sup(F)=sup(|F|)
Recall that if F=F then we know sup(F)=sup(|F|)
Supremum of the Sum of Two Functions is Less than or Equal to the Sum of the Supremums
Suppose that f,g:A then sup({f(x)+g(x):xA})sup({f(x):xA})+sup({g(x):xA})

Let α=sup({f(x):xA}),β=sup({g(x):xA}) and γ=sup({f(x)+g(x):xA}).

We know that for any xA we have that f(x)α and g(x)β therefore we have that f(x)+g(x)α+β, this proves that α+β is an upper bound to the set {f(x)+g(x):xA} therefore we have sup({f(x)+g(x):xA})sup({f(x):xA})+sup({f(x):xA}) As γ is the least upper bound.

Supremum of the Sum of two Functions is the Same as the Sum of the Supremums if the Variables are Independent
Suppose that f,g:A then sup({f(x)+g(y):x,yA})=sup({f(x):xA})+sup({g(x):xA})

Let α=sup({f(x):xA}),β=sup({g(x):xA}) and γ=sup({f(x)+g(y):x,yA}).

For any xA we have that f(x)α and for any yA we have g(x)β thus we have f(x)+g(y)α+β so that α+β is an upper bound of the set γ is supping over there fore we have that γα+β

... [TODO] not sure how to prove the other direction ...

Bound on the Upper minus Lower Sum for The Same Partition
Suppose that f:[a,b] is bounded, and let P={x1,,xn+1} be a partition of [a,b] then we have that U(f,P)L(f,P)2f(ba)
U(f,P)L(f,P)amp;=i=1n(supx[xi,xi+1]f(x)inf(x[xi,xi+1]f(x)))Δi=1i=1n(supx[xi,xi+1]f(x)+sup(x[xi,xi+1]f(x)))Δi=2i=1n(supx,y[xi,xi+1](f(x)f(y)))Δi=3i=1n(supx,y[xi,xi+1](|f(x)f(y)|))Δi4i=1n(2supx,y[xi,xi+1](|f(x)|))Δi5i=1n(2f)Δiamp;=2fi=1nΔiamp;=62f(ba) Where we've used
Piecewise Continuous
A function f:[a,b] is called piecewise continuous if for every [c,d][a,b] if it only has a finite number of discontinuities all of which are jump discontinuities
Riemann Integrable on Two Parts of the Interval Means Riemann Integrable on the Whole Interval
Suppose that f:[a,b] and let x[a,b] then if f restricted to [a,x] and f restricted to [x,b] (which we denote as g1,g2) are both riemann integrable then so is f on [a,b] and then abf(x)dx=axf(x)dx+xbf(x)dx
Let ϵ+ using bullet point 2 from the RMI characterization we know that there are partitions P1,P2 of [a,x],[x,b] such that (U(g1,P1)L(g1,P1))+(U(g2,P2)L(g2,P2))<ϵ Then note that U(g1,P1)+U(g2,P2)=U(f,P1P2) which also holds for the lower sums, and therefore from the previous inequality we have U(f,P1P2)L(f,P1P2)<ϵ therefore by (2) of the RMI characterization f is Riemann integrable.
Every Piecewise Continuous Function is Riemann Integrable

Since f is piecewise continuous, then suppose its jump discontinuity points are given by D={d1,d2,dn}. We will show this by using strong induction on the number of discontinuities within [a,b].

Note that one can manage to prove that if f is RMI over [a,b] with 1ik jump discontinuities, then it is RMI over [a,b] if it has k+1 discontinuities. This can be done because we can break [a,b] into two intervals I1,I2 such that I1I2=[a,b], where I1,I2 each have a number of discontinuities within the range [1,,k] so that fI1 and fI2 are both integrable on their respective intervals, so that f is RMI on [a,b].

Therefore all that remains is the base case, which is that f is RMI if there is exactly 1 jump discontinuity of f on the set [a,b], let d be this discontinuity, since f is continuous on L:=[a,dδ] and R:=[d+δ,b] for any small δ, then we know that f is RMI when restricted to these intervals.

Let ϵ+ by (2) of the RMI characterization we obtain PL,PR which are partitions of L,R respectively (note that PLPR= such that U(fL,PL)L(fL,PL)<ϵ3 and U(fR,PR)L(fR,PR)<ϵ3

Finally Consider PLPR which is a partition of [a,b] which contains the section [dδ,d+δ] along with the left and right partitions. Focusing on this section we can see it will contribute (supx[dδ,d+δ]f(x)supy[dδ,d+δ]f(y))2δ to U(f,PLPR)L(f,PLPR), now:

U(f,PLPR)L(f,PLPR)amp;=(U(fL,PL)+(supx[dδ,d+δ]f(x))2δ+U(fR,PR))(L(fL,PL)+(infx[dδ,d+δ]f(x))2δ+L(fR,PR))amp;=(U(fL,PL)L(fL,PL))+(supx[dδ,d+δ]f(x)infx[dδ,d+δ]f(x))2δ+(U(fR,PR)L(fL,PL))amp;<2ϵ3+supx,y[dδ,d+δ](|f(x)f(y)|)2δamp;2ϵ3+2supx[dδ,d+δ]|f(x)|amp;2ϵ3+2f2δamp;=2ϵ3+4fδ Thus a selection of δ=ϵ4f3 shows that U(f,PLPR)L(f,PLPR)<ϵ as needed.