Differentiation of integrals

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

\lim_{r \to 0} \frac1{\mu \big( B_{r} (x) \big)} \int_{B_{r} (x)} f(y) \, \mathrm{d} \mu(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.

Theorems on the differentiation of integrals

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 \to 0} \frac1{\lambda^{n} \big( B_{r} (x) \big)} \int_{B_{r} (x)} f(y) \, \mathrm{d} \lambda^{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 \to 0} \frac1{\mu \big( B_{r} (x) \big)} \int_{B_{r} (x)} f(y) \, \mathrm{d} \mu (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:

\lim_{r \to 0} \frac{\gamma \big( M \cap B_{r} (x) \big)}{\gamma \big( B_{r} (x) \big)} = 1.
\lim_{r \to 0} \inf \left\{ \left. \frac1{\gamma \big( B_{s} (x) \big)} \int_{B_{s} (x)} f(y) \, \mathrm{d} \gamma(y) \right| x \in H, 0 < s < r \right\} = + \infty.

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

\langle Sx, y \rangle = \int_{H} \langle x, z \rangle \langle y, z \rangle \, \mathrm{d} \gamma(z),

or, for some countable orthonormal basis (ei)iN of H,

Sx = \sum_{i \in \mathbf{N}} \sigma_{i}^{2} \langle x, e_{i} \rangle e_{i}.

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

\sigma_{i + 1}^{2} \leq q \sigma_{i}^{2},

then, for all f  L1(H, γ; R),

\frac1{\mu \big( B_{r} (x) \big)} \int_{B_{r} (x)} f(y) \, \mathrm{d} \mu(y) \xrightarrow[r \to 0]{\gamma} f(x),

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

\sigma_{i + 1}^{2} \leq \frac{\sigma_{i}^{2}}{i^{\alpha}}

for some α > 5  2, then

\frac1{\mu \big( B_{r} (x) \big)} \int_{B_{r} (x)} f(y) \, \mathrm{d} \mu(y) \xrightarrow[r \to 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 \to 0} \frac1{\gamma \big( B_{r} (x) \big)} \int_{B_{r} (x)} f(y) \, \mathrm{d} \gamma(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.

See also

References

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