Girish Mahajan (Editor)

Differentiation of integrals

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In mathematics, the problem of differentiation of integrals is that of determining under what circumstances the mean value integral of a suitable function on a small neighbourhood of a point approximates the value of the function at that point. More formally, given a space X with a measure μ and a metric d, one asks for what functions f : X → R does

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lim r → 0 1 μ ( B r ( x ) ) ∫ B r ( x ) f ( y ) d μ ( y ) = f ( x )

for all (or at least μ-almost all) x ∈ X? (Here, as in the rest of the article, Br(x) denotes the open ball in X with d-radius r and centre x.) This is a natural question to ask, especially in view of the heuristic construction of the Riemann integral, in which it is almost implicit that f(x) is a "good representative" for the values of f near x.

Lebesgue measure

One result on the differentiation of integrals is the Lebesgue differentiation theorem, as proved by Henri Lebesgue in 1910. Consider n-dimensional Lebesgue measure λn on n-dimensional Euclidean space Rn. Then, for any locally integrable function f : Rn → R, one has

lim r → 0 1 λ n ( B r ( x ) ) ∫ B r ( x ) f ( y ) d λ n ( y ) = f ( x )

for λn-almost all points x ∈ Rn. It is important to note, however, that the measure zero set of "bad" points depends on the function f.

Borel measures on Rn

The result for Lebesgue measure turns out to be a special case of the following result, which is based on the Besicovitch covering theorem: if μ is any locally finite Borel measure on Rn and f : Rn → R is locally integrable with respect to μ, then

lim r → 0 1 μ ( B r ( x ) ) ∫ B r ( x ) f ( y ) d μ ( y ) = f ( x )

for μ-almost all points x ∈ Rn.

Gaussian measures

The problem of the differentiation of integrals is much harder in an infinite-dimensional setting. Consider a separable Hilbert space (H, 〈 , 〉) equipped with a Gaussian measure γ. As stated in the article on the Vitali covering theorem, the Vitali covering theorem fails for Gaussian measures on infinite-dimensional Hilbert spaces. Two results of David Preiss (1981 and 1983) show the kind of difficulties that one can expect to encounter in this setting:

  • There is a Gaussian measure γ on a separable Hilbert space H and a Borel set M ⊆ H so that, for γ-almost all x ∈ H,
  • There is a Gaussian measure γ on a separable Hilbert space H and a function f ∈ L1(H, γ; R) such that

    • However, there is some hope if one has good control over the covariance of γ. Let the covariance operator of γ be S : H → H given by

    ⟨ S x , y ⟩ = ∫ H ⟨ x , z ⟩ ⟨ y , z ⟩ d γ ( z ) ,

    or, for some countable orthonormal basis (ei)i∈N of H,

    S x = ∑ i ∈ N σ i 2 ⟨ x , e i ⟩ e i .

    In 1981, Preiss and Jaroslav Tišer showed that if there exists a constant 0 < q < 1 such that

    σ i + 1 2 ≤ q σ i 2 ,

    then, for all f ∈ L1(H, γ; R),

    1 μ ( B r ( x ) ) ∫ B r ( x ) f ( y ) d μ ( y ) → r → 0 γ f ( x ) ,

    where the convergence is convergence in measure with respect to γ. In 1988, Tišer showed that if

    σ i + 1 2 ≤ σ i 2 i α

    for some α > 5 ⁄ 2, then

    1 μ ( B r ( x ) ) ∫ B r ( x ) f ( y ) d μ ( y ) → r → 0 f ( x ) ,

    for γ-almost all x and all f ∈ Lp(H, γ; R), p > 1.

    As of 2007, it is still an open question whether there exists an infinite-dimensional Gaussian measure γ on a separable Hilbert space H so that, for all f ∈ L1(H, γ; R),

    lim r → 0 1 γ ( B r ( x ) ) ∫ B r ( x ) f ( y ) d γ ( y ) = f ( x )

    for γ-almost all x ∈ H. However, it is conjectured that no such measure exists, since the σi would have to decay very rapidly.

    References

    Differentiation of integrals Wikipedia
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