counter-example of Fubini’s theorem for
the Lebesgue integral∗
rmilson†
2013-03-22 1:31:33
The following observation demonstrates the necessity of the integrability
assumption in Fubini’s theorem. Let
Q = {(x, y) ∈ R2 : x ≥ 0, y ≥ 0}
denote the upper, right quadrant. Let R ⊂ Q be the region in the quadrant
bounded by the lines y = x, y = x − 1, and let let S ⊂ Q be a similar region,
but this time bounded by the lines y = x − 1, y = x − 2. Let
f = χS − χR ,
where χ denotes a characteristic function.
Observe that the Lebesgue measure of R and of S is infinite. Hence, f is
not a Lebesgue-integrable function. However for every x ≥ 0 the function
Z ∞
g(x) =
f (x, y) dy
0
is integrable. Indeed,

 −x
x−2
g(x) =

0
for 0 ≤ x ≤ 1,
for 1 ≤ x ≤ 2,
for x ≥ 2.
Similarly, for y ≥ 0, the function
Z
∞
h(y) =
f (x, y) dx
0
is integrable. Indeed,
h(y) = 0,
y ≥ 0.
∗ hCounterexampleOfFubinisTheoremForTheLebesgueIntegrali
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1
Hence, the values of the iterated integrals
Z ∞
g(x) dx = −1,
0
Z
∞
h(y) dy = 0,
0
are finite, but do not agree. This does not contradict Fubini’s theorem because
the value of the planar Lebesgue integral
Z
f (x, y) dµ(x, y),
Q
where µ(x, y) is the planar Lebesgue measure, is not defined.
2