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The graph of a continuous function R→R is Lebesgue null in R2

Example

Assume the Axiom of Countable Choice and let f:R→R be continuous. Then its graph

Γf:={ (x,f(x)):x∈R }

is Lebesgue null in R2.

Facts & Assumptions

Given: The Axiom of Countable Choice and a continuous function f:R→R.

[L1]

A subset of Rm has Lebesgue outer measure zero if and only if it is null in the sense of countable closed-cube covers (A subset of Rm has Lebesgue outer measure zero if and only if it is null in the sense of countable closed-cube covers).

[L2]

Assuming countable choice, a box in Rn with parameters ai≤bi is Lebesgue measurable of measure ∏i<n(bi−ai) (A box in Rn with parameters ai≤bi is Lebesgue measurable of measure ∏i<n(bi−ai), whichever of its faces are included).

[F1]

Let μ be a measure and let (Ek)k∈N be measurable. Then μ(⋃kEk)≤∑kμ(Ek) (Finite and countable subadditivity of measures).

[L3]

A continuous real function on a compact subset of R is uniformly continuous (Heine-Cantor in R: a continuous real function on a compact subset of R is uniformly continuous).

[F4]

For every real δ>0 there is a natural number N≥1 with 1/N<δ (For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε).

[F5]

If ∣r∣<1 then the series ∑rk converges; in particular ∑n=0∞2−n=2 (For ∣r∣<1, ∑k≥0rk=1/(1−r), and for ∣r∣≥1 the series diverges).

Verification

technique · direct
1.1F2F3L3F4

Fix a real η>0. For each integer m, the interval [m,m+1] is compact by [F3], so [L3] gives a real δm>0 such that x,y∈[m,m+1] and ∣x−y∣<δm imply ∣f(x)−f(y)∣<η2−∣m∣−3. By [F4] choose a natural number Nm≥1 with 1/Nm<δm, and put tj:=m+j/Nm for 0≤j≤Nm. Then for every j with 1≤j≤Nm and every x∈[tj−1,tj] one has ∣x−tj−1∣≤1/Nm<δm, hence ∣f(x)−f(tj−1)∣<η2−∣m∣−3.

2.1step 1.1L2algebra

For each such subinterval, the corresponding graph piece lies in the closed box [tj−1,tj]×[f(tj−1)−η2−∣m∣−3, f(tj−1)+η2−∣m∣−3], whose width is tj−tj−1 and whose height is η2−∣m∣−2. Therefore [L2] gives a box cover of Γf∩([m,m+1]×R) whose total area is ∑j=1Nm(tj−tj−1) η2−∣m∣−2=η2−∣m∣−2.

3.1step 2.1L1F1F5algebra∎

Summing these covers over all integers m and using [F1], the whole graph is covered by countably many closed boxes with total area at most η∑m∈Z2−∣m∣−2=η(2−2+2∑n=1∞2−n−2)=3η/4<η, the geometric-series identity coming from [F5]. Since η>0 was arbitrary, [L1] gives that Γf is Lebesgue null.

Depends on

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