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∫03⌊x⌋=3: the floor function is nondecreasing, hence integrable, and the integral is computed from the uniform partitions

Example

Let g:[0,3]→R be g(x)=⌊x⌋, the integer part (Integer part: for every real x there is exactly one integer m with m≤x<m+1). Then g is nondecreasing (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of R, with the dictionary to monotone sequences), hence Riemann integrable on [0,3] (A monotone function on [a,b] is Riemann integrable: for the uniform partition into N parts the upper minus lower sum telescopes to ∣f(b)−f(a)∣ (b−a)/ι(N)), and

∫03⌊x⌋ dx  =  3.

g is discontinuous at x=1, x=2 and x=3 and continuous elsewhere on [0,3] (Discontinuity of f at a point of its domain, and its classification: removable discontinuity, jump discontinuity and essential discontinuity, equivalently Rudin's discontinuities of the first and of the second kind), so this is an integrable function with genuine jumps, not a continuous one in disguise; the value 3 is 0⋅1+1⋅1+2⋅1, the three constant pieces weighted by their lengths.

The computation below uses the uniform partition into N=3M parts, M≥1, for which the lower sum is exactly 3 at every M and the upper sum is 3+3/M. So the lower sums do not merely approach the integral, they attain it.

Facts & Assumptions

Given: g:[0,3]→R with g(x)=⌊x⌋; a natural M≥1; N:=3M; and the uniform partition UN=(N,t) of [0,3] with ti=3ι(i)/ι(N)=ι(i)/ι(M) for i≤N and lengths Δi=3/ι(N)=1/ι(M).

[L1]

For every real x there is exactly one integer ⌊x⌋ with ⌊x⌋≤x<⌊x⌋+1 (Integer part: for every real x there is exactly one integer m with m≤x<m+1).

[L2]

g is nondecreasing: for x≤y, ⌊x⌋≤x≤y<⌊y⌋+1, and ⌊x⌋, ⌊y⌋ are integers, so ⌊x⌋≤⌊y⌋, no integer lying strictly between ⌊y⌋ and ⌊y⌋+1 (Integer part: for every real x there is exactly one integer m with m≤x<m+1, Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of R, with the dictionary to monotone sequences).

[L6]

Finite sums: splitting a sum over i<3M into the three blocks i<M, M≤i<2M and 2M≤i<3M; scaling; and ∑i<Mc=ι(M)c (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L7]

For i∈N: ⌊ι(i)/ι(M)⌋=0 when i<M, =1 when M≤i<2M, and =2 when 2M≤i<3M; and ⌊ι(j)/ι(M)⌋=3 for j=3M. Each case is [L1] applied to the displayed inequalities 0≤ι(i)/ι(M)<1, 1≤ι(i)/ι(M)<2, 2≤ι(i)/ι(M)<3 and ι(3M)/ι(M)=3, which follow from ι being strictly increasing and additive (Canonical naturals are positive and strictly increasing, The canonical natural ι(n)=n⋅1F of a field, Order is preserved by adding a constant and by adding inequalities). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used here follow by adjoining the equality case, in which the two sides coincide.

[L9]

Ordered-field arithmetic and the absolute value: adding a constant and multiplying by a positive quantity preserve an inequality; the order is total and transitive (Basic properties of the absolute value, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Verification

technique · direct
1.1

g is nondecreasing by [L2], and 0<3, so g is Riemann integrable on [0,3] by [L3]; write J:=∫03g.

givenL2L3
1.2

By [L5] and [L7], for i<N=3M the lower value is mi=g(ti)=⌊ι(i)/ι(M)⌋, which is 0 for i<M, 1 for M≤i<2M and 2 for 2M≤i<3M; and the upper value is Mi=g(ti+1)=⌊ι(i+1)/ι(M)⌋, which is 0 for i+1<M, 1 for M≤i+1<2M, 2 for 2M≤i+1<3M and 3 for i+1=3M.

givenL1L4L5L7
2.1

By [L6] and step 1.2, L(g,UN)=1ι(M)(∑i<M0+∑M≤i<2M1+∑2M≤i<3M2)=1ι(M)(0+ι(M)+2ι(M))=3.

step 1.2L4L5L6L9
2.2

By [L6] and step 1.2, the upper values run over j=i+1 from 1 to 3M, giving ι(M)−1 indices with value 0, then ι(M) with value 1, then ι(M) with value 2, and the single index j=3M with value 3; hence U(g,UN)=1ι(M)(0+ι(M)+2ι(M)+3)=3+3ι(M).

step 1.2L4L5L6L9
3.1

By [L5], 3=L(g,UN)≤∫03‾g≤J≤∫03‾g≤U(g,UN)=3+3/ι(M) for every natural M≥1.

step 1.1step 2.1step 2.2L5
4.1

Hence 0≤J−3≤3/ι(M) for every M≥1. If J>3 then (J−3)⋅3−1>0 and [L8] supplies M≥1 with 1/ι(M)<(J−3)⋅3−1, that is 3/ι(M)<J−3, contradicting step 3.1. So J=3, that is ∫03⌊x⌋ dx=3.

step 3.1L8L9∎

Remarks

Depends on

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