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 → 0 1 μ ( B r ( x ) ) ∫ B r ( x ) f ( y ) d μ ( y ) = f ( x ) {\displaystyle \lim _{r\to 0}{\frac {1}{\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 → 0 1 λ n ( B r ( x ) ) ∫ B r ( x ) f ( y ) d λ n ( y ) = f ( x ) {\displaystyle \lim _{r\to 0}{\frac {1}{\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. 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 ) {\displaystyle \lim _{r\to 0}{\frac {1}{\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:
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, lim r → 0 γ ( M ∩ B r ( x ) ) γ ( B r ( x ) ) = 1. {\displaystyle \lim _{r\to 0}{\frac {\gamma {\big (}M\cap B_{r}(x){\big )}}{\gamma {\big (}B_{r}(x){\big )}}}=1.}
There is a Gaussian measure γ on a separable Hilbert space H and a function f ∈ L1(H, γ; R) such that lim r → 0 inf { 1 γ ( B s ( x ) ) ∫ B s ( x ) f ( y ) d γ ( y ) | x ∈ H , 0 < s < r } = + ∞ . {\displaystyle \lim _{r\to 0}\inf \left\{\left.{\frac {1}{\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
⟨ S x , y ⟩ = ∫ H ⟨ x , z ⟩ ⟨ y , z ⟩ d γ ( z ) , {\displaystyle \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)i∈N of H,
S x = ∑ i ∈ N σ i 2 ⟨ x , e i ⟩ e i . {\displaystyle 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
σ i + 1 2 ≤ q σ i 2 , {\displaystyle \sigma _{i+1}^{2}\leq q\sigma _{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 ) , {\displaystyle {\frac {1}{\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
σ i + 1 2 ≤ σ i 2 i α {\displaystyle \sigma _{i+1}^{2}\leq {\frac {\sigma _{i}^{2}}{i^{\alpha }}}}
for some α > 5 ⁄ 2, then
1 μ ( B r ( x ) ) ∫ B r ( x ) f ( y ) d μ ( y ) → r → 0 f ( x ) , {\displaystyle {\frac {1}{\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 → 0 1 γ ( B r ( x ) ) ∫ B r ( x ) f ( y ) d γ ( y ) = f ( x ) {\displaystyle \lim _{r\to 0}{\frac {1}{\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 Differentiation rules – Rules for computing derivatives of functions Leibniz integral rule – Differentiation under the integral sign formula Reynolds transport theorem – 3D generalization of the Leibniz integral rule
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