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03x=3\int_0^3 \lfloor x \rfloor = 3: the floor function is nondecreasing, hence integrable, and the integral is computed from the uniform partitions

Example

Let g:[0,3]Rg : [0,3] \to \mathbb{R} be g(x)=xg(x) = \lfloor x \rfloor, the integer part (Integer part: for every real xx there is exactly one integer mm with mx<m+1m \le x < m + 1). Then gg is nondecreasing (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of R\mathbb{R}, with the dictionary to monotone sequences), hence Riemann integrable on [0,3][0,3] (A monotone function on [a,b][a,b] is Riemann integrable: for the uniform partition into NN parts the upper minus lower sum telescopes to f(b)f(a)(ba)/ι(N)|f(b) - f(a)|\,(b-a)/\iota(N)), and

03xdx  =  3.\int_0^3 \lfloor x \rfloor \,\mathrm{d}x \;=\; 3 .

gg is discontinuous at x=1x = 1, x=2x = 2 and x=3x = 3 and continuous elsewhere on [0,3][0,3] (Discontinuity of ff 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 33 is 01+11+210\cdot 1 + 1\cdot 1 + 2\cdot 1, the three constant pieces weighted by their lengths.

The computation below uses the uniform partition into N=3MN = 3M parts, M1M \ge 1, for which the lower sum is exactly 33 at every MM and the upper sum is 3+3/M3 + 3/M. So the lower sums do not merely approach the integral, they attain it.

Facts & Assumptions

Given: g:[0,3]Rg : [0,3] \to \mathbb{R} with g(x)=xg(x) = \lfloor x \rfloor; a natural M1M \ge 1; N:=3MN := 3M; and the uniform partition UN=(N,t)U_N = (N,t) of [0,3][0,3] with ti=3ι(i)/ι(N)=ι(i)/ι(M)t_i = 3\iota(i)/\iota(N) = \iota(i)/\iota(M) for iNi \le N and lengths Δi=3/ι(N)=1/ι(M)\Delta_i = 3/\iota(N) = 1/\iota(M).

[L1]

For every real xx there is exactly one integer x\lfloor x \rfloor with xx<x+1\lfloor x \rfloor \le x < \lfloor x \rfloor + 1 (Integer part: for every real xx there is exactly one integer mm with mx<m+1m \le x < m + 1).

[L2]

gg is nondecreasing: for xyx \le y, xxy<y+1\lfloor x \rfloor \le x \le y < \lfloor y \rfloor + 1, and x\lfloor x \rfloor, y\lfloor y \rfloor are integers, so xy\lfloor x \rfloor \le \lfloor y \rfloor, no integer lying strictly between y\lfloor y \rfloor and y+1\lfloor y \rfloor + 1 (Integer part: for every real xx there is exactly one integer mm with mx<m+1m \le x < m + 1, Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of R\mathbb{R}, with the dictionary to monotone sequences).

[L5]

For a nondecreasing gg and a subinterval Ii=[ti,ti+1]I_i = [t_i,t_{i+1}]: mi=g(ti)m_i = g(t_i) and Mi=g(ti+1)M_i = g(t_{i+1}), both attained; L(g,P)=i<NmiΔiL(g,P) = \sum_{i<N}m_i\Delta_i, U(g,P)=i<NMiΔiU(g,P) = \sum_{i<N}M_i\Delta_i, and L(g,P)03g03gU(g,P)L(g,P) \le \underline{\int_0^3}g \le \overline{\int_0^3}g \le U(g,P) (For bounded ff on [a,b][a,b] and a partition PP: the infimum mim_i and supremum MiM_i of ff on the ii-th subinterval, and the lower and upper Darboux sums L(f,P)=imiΔiL(f,P) = \sum_i m_i \Delta_i and U(f,P)=iMiΔiU(f,P) = \sum_i M_i \Delta_i, The lower and upper Darboux integrals of a bounded ff on [a,b][a,b] as supPL(f,P)\sup_P L(f,P) and infPU(f,P)\inf_P U(f,P), Darboux integrability as their equality, and the notation abf\int_a^b f, Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L6]

Finite sums: splitting a sum over i<3Mi < 3M into the three blocks i<Mi < M, Mi<2MM \le i < 2M and 2Mi<3M2M \le i < 3M; scaling; and i<Mc=ι(M)c\sum_{i<M}c = \iota(M)c (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L7]

For iNi \in \mathbb{N}: ι(i)/ι(M)=0\lfloor \iota(i)/\iota(M) \rfloor = 0 when i<Mi < M, =1= 1 when Mi<2MM \le i < 2M, and =2= 2 when 2Mi<3M2M \le i < 3M; and ι(j)/ι(M)=3\lfloor \iota(j)/\iota(M)\rfloor = 3 for j=3Mj = 3M. Each case is [L1] applied to the displayed inequalities 0ι(i)/ι(M)<10 \le \iota(i)/\iota(M) < 1, 1ι(i)/ι(M)<21 \le \iota(i)/\iota(M) < 2, 2ι(i)/ι(M)<32 \le \iota(i)/\iota(M) < 3 and ι(3M)/ι(M)=3\iota(3M)/\iota(M) = 3, which follow from ι\iota being strictly increasing and additive (Canonical naturals are positive and strictly increasing, The canonical natural ι(n)=n1F\iota(n) = n \cdot 1_F 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

gg is nondecreasing by [L2], and 0<30 < 3, so gg is Riemann integrable on [0,3][0,3] by [L3]; write J:=03gJ := \int_0^3 g.

givenL2L3
1.2

By [L5] and [L7], for i<N=3Mi < N = 3M the lower value is mi=g(ti)=ι(i)/ι(M)m_i = g(t_i) = \lfloor \iota(i)/\iota(M)\rfloor, which is 00 for i<Mi < M, 11 for Mi<2MM \le i < 2M and 22 for 2Mi<3M2M \le i < 3M; and the upper value is Mi=g(ti+1)=ι(i+1)/ι(M)M_i = g(t_{i+1}) = \lfloor \iota(i+1)/\iota(M)\rfloor, which is 00 for i+1<Mi+1 < M, 11 for Mi+1<2MM \le i+1 < 2M, 22 for 2Mi+1<3M2M \le i+1 < 3M and 33 for i+1=3Mi+1 = 3M.

givenL1L4L5L7
2.1

By [L6] and step 1.2, L(g,UN)=1ι(M)(i<M0+Mi<2M1+2Mi<3M2)=1ι(M)(0+ι(M)+2ι(M))=3L(g,U_N) = \frac{1}{\iota(M)}\Bigl(\sum_{i<M}0 + \sum_{M \le i < 2M}1 + \sum_{2M \le i < 3M}2\Bigr) = \frac{1}{\iota(M)}\bigl(0 + \iota(M) + 2\iota(M)\bigr) = 3.

step 1.2L4L5L6L9
2.2

By [L6] and step 1.2, the upper values run over j=i+1j = i+1 from 11 to 3M3M, giving ι(M)1\iota(M)-1 indices with value 00, then ι(M)\iota(M) with value 11, then ι(M)\iota(M) with value 22, and the single index j=3Mj = 3M with value 33; hence U(g,UN)=1ι(M)(0+ι(M)+2ι(M)+3)=3+3ι(M)U(g,U_N) = \frac{1}{\iota(M)}\bigl(0 + \iota(M) + 2\iota(M) + 3\bigr) = 3 + \frac{3}{\iota(M)}.

step 1.2L4L5L6L9
3.1

By [L5], 3=L(g,UN)03gJ03gU(g,UN)=3+3/ι(M)3 = L(g,U_N) \le \underline{\int_0^3} g \le J \le \overline{\int_0^3} g \le U(g,U_N) = 3 + 3/\iota(M) for every natural M1M \ge 1.

step 1.1step 2.1step 2.2L5
4.1

Hence 0J33/ι(M)0 \le J - 3 \le 3/\iota(M) for every M1M \ge 1. If J>3J > 3 then (J3)31>0(J-3)\cdot 3^{-1} > 0 and [L8] supplies M1M \ge 1 with 1/ι(M)<(J3)311/\iota(M) < (J-3)\cdot 3^{-1}, that is 3/ι(M)<J33/\iota(M) < J - 3, contradicting step 3.1. So J=3J = 3, that is 03xdx=3\int_0^3 \lfloor x \rfloor\,\mathrm{d}x = 3.

step 3.1L8L9

Remarks

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