Here is a consequence of Lebesgue dominated convergence theorem on differentiation under integral sign.

Function $f(x, t)$ is differentiable at $x_0$ for almost all $t \in A$, and $t \to f(x, t)$ is integrable. Moreover, there exist an integrable function $g(t)$ such that $|(\partial{f}/\partial{x})(x,t)| \le g(t)$. Then we have $$\frac{\mathrm{d}}{\mathrm{d}x} \left( \int_A f(x,t) ~ \mathrm{d}t \right)\bigg|_{x=x_0} = \int \frac{\partial{f}}{\partial{x}}(x_0,t) ~ \mathrm{d}t. $$

Of course, here the word "integrable" means "Lebesgue integrable". But, if we read the word "integrable" as "improper Riemann integrable" and add an assumption that the partial derivative is continuous, then I think the statement is still true. Weierstrass M-test for integrals guarantees uniform convergence of the improper integral and we can interchange the differentiaion and integration. If $f$ is integrable in both of (improper) Riemann and Lebesgue sense, the only gain of interpreting the integral as Lebesgue one is gettig rid of the continuity condition of the partial derivative. Is it right?

  • $\begingroup$ In your right hand side is it supposed to be df/dx(x,t)dt evaluated at x = x0 and not df/dx(x0,t)? $\endgroup$ Commented Dec 5, 2012 at 5:07

1 Answer 1


What you write is more or less true. First note that it is not enough to assume that $f(x,t)$ is differentiable at $x_0$ (for (almost) all $t$), we need that there is a neighorhood $(x_0 - \varepsilon, x_0 + \varepsilon) =:I$ such that $f(\cdot, t)$ is differentiable in $x$ for all $x \in I$ and (almost) all $t \in A$, where the "exceptional null set" must not depend on $x$.

The assumption $|\partial f/\partial x (x,t)| \leq g(x)$ only makes sense in this setting, otherwise we could only require $x_0$ instead of $x$ and the proof also needs this assumption (see below).

Now, let $h_n \to 0$ with $h_n \neq 0$ for all $n$ and w.l.o.g. with $x_0 + h_n \in I$ for all $n$. Let $$ F(x) := \int_A f(x,t) \, dt \text{ for } x \in I,\\ G(x) := \int_A \frac{\partial f}{\partial x}(x,t) \, dt \text{ for } x \in I. $$ Here, the integral defining $F$ is interpreted as an improper Riemann integral (it is possible that this integral does not exist as a Lebesgue integral), but the integral defining $G$ is interpreted as a Lebesgue integral. If $t \mapsto \frac{\partial f}{\partial x}$ is Riemann integrable over every compact subinterval of $A$ (for example, if it is continuous in $t$), then it can also be interpreted as an improper Riemann integral, at least if we assume that the dominating function $g$ is (improper) Riemann integrable.

Observe that the assumptions guarantee that $F,G$ are well-defined. Measurability of $t \mapsto \frac{\partial f}{\partial x}(x,t)$ for each $x \in I$ is implied by the pointwise convergence $$\frac{\partial f}{\partial x}(x,t) = \lim_n \frac{f(x+h_n, t) - f(x,t)}{h_n}, $$ which will also be employed below.

We now have \begin{eqnarray*} \left|\frac{F\left(x_{0}+h_{n}\right)-F\left(x_{0}\right)}{h_{n}}-G\left(x_{0}\right)\right| & = & \left|\int_{A}\frac{f\left(x_{0}+h_{n},t\right)-f\left(x_{0},t\right)}{h_{n}}-\frac{\partial f}{\partial x}\left(x_{0},t\right)\,{\rm d}t\right|\\ & \leq & \int_{A}\left|\frac{f\left(x_{0}+h_{n},t\right)-f\left(x_{0},t\right)}{h_{n}}-\frac{\partial f}{\partial x}\left(x_{0},t\right)\right|\,{\rm d}t. \end{eqnarray*} Note that the integrands are measurable and that for (almost) all $t \in A$, the mean value theorem yields some $\xi_{n,t} \in I$ (even between $x_0$ and $x_0 +h_n$) such that $$ \left|\frac{f\left(x_{0}+h_{n},t\right)-f\left(x_{0},t\right)}{h_{n}}\right|=\left|\frac{\partial f}{\partial x}\left(\xi_{n,t},t\right)\right|\leq g\left(t\right). $$ This shows that the integrands in the integrals are Lebesgue-integrable and dominated by an integrable function. As the integrand in the last integral converges pointwise to $0$ for $n \to \infty$, the dominated convergence theorem yields $$ \left|\frac{F\left(x_{0}+h_{n}\right)-F\left(x_{0}\right)}{h_{n}}-G\left(x_{0}\right)\right| \to 0, $$ so that $F$ is differentiable (in $x_0$) with the expected derivative.

  • $\begingroup$ If you have time can you take a look at this notes at page 130,exercise 34?...Is the condition $(ii)$ correct? i believe that we must have integrability for all $t \in (a,b)$ What do you think? fourier.math.uoc.gr/~papadim/measure_theory.pdf $\endgroup$ Commented Dec 17, 2019 at 0:29
  • $\begingroup$ @Marios: I think this is correct as stated. You should be able to use the mean value theorem and the bound on the derivative to get from the integrability for one $t_0$ that it is actually integrable for all $t$. $\endgroup$
    – PhoemueX
    Commented Dec 17, 2019 at 9:54

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