The problem is the following:

Let $a,b,c,d \in \mathbb R$ be given such that $a<b$ and $c<d$. Suppose $f : [a,b] \times [c,d] \to \mathbb R$ is a function such that $\partial_1 f: [a,b] \times [c,d] \to \mathbb R$ exists and is (Lebesgue-)integrable. Show that for $t\in [a,b]$ we have $$\frac{\mathrm d}{\mathrm dt}\int_c^d f(t,x) \, \mathrm dx = \int_c^d \partial_1 f(t,x)\, \mathrm dx$$

I have tried the following: By the fundamental theorem of calculus, we have

$$f(t',x) - f(t,x) = \int_{t}^{t'} \partial_1f(s,x)\, \mathrm ds$$

It follows that

$$ \begin{align} \int_c^d f(t',x)-f(t,x) \, \mathrm dx &= \int_c^d \int_{t}^{t'} \partial_1f(s,x)\, \mathrm ds \, \mathrm dx \\ &= \int_{t}^{t'} \int_c^d \partial_1f(s,x)\, \mathrm dx \, \mathrm ds \end{align} $$

so that for almost all $t$, we have

$$ \begin{align} \frac{\mathrm d}{\mathrm dt}\int_c^d f(t,x) \, \mathrm dx &= \lim_{t'\to t} \frac{1}{t'-t}\int_c^d f(t',x)-f(t,x) \, \mathrm dx \\ &= \lim_{t'\to t} \frac{1}{t'-t}\int_{t}^{t'} \int_c^d \partial_1f(s,x)\, \mathrm dx \, \mathrm ds \\ &= \int_c^d \partial_1f(t,x)\, \mathrm dx \end{align} $$

since $t \mapsto \int_c^d \partial_1f(t,x)\, \mathrm dx$ is an $L^1$ function (hence almost every point $t$ is a Lebesgue-point).

Unfortunately equality for almost every $t$ is not good enough in this case.

I have been thinking about this for quite a while now, but I cannot figure out the reason for why every point $t\in [a,b]$ has to be a Lebesgue point of

$$t\mapsto \int_c^d \partial_1f(t,x)\, \mathrm dx$$

I have also been thinking about possible counterexamples, but couldn't really come up with one. (At least not if one defines $\frac{\mathrm d}{\mathrm dt} \infty = 0$ for the constant function $\infty$...)

Some help would be appreciated, thanks!


You need $\partial_1f(t,x)$ be dominated by an integrable function, i.e., you need $|\partial_1f(t,x)|\le g(x)$ for every $x,t$ for you to take the limit $t'\to t$ inside the integral for every $t$.

Just $\partial_1f(t,x)$ being integrable over $[c,d]$ for every fixed $t\in [a,b]$ isn't sufficient.


Let me clarify things here. The problem is that you can not straightaway write $$\frac{\mathrm d}{\mathrm dt}\int_c^d f(t,x) \, \mathrm dx = \lim_{t'\to t} \frac{1}{t'-t}\int_c^d f(t',x)-f(t,x) \, \mathrm dx $$ unless you have already shown that $\int_c^d f(t,x) \, \mathrm dx$ is differentiable w.r.to $t$.

So, let $t_n$ be a sequence converging to $t$ and let $$g_n(x)=\frac{f(t_n,x)-f(t,x)}{t_n-t}$$ Then $\lim g_n(x)=\partial_1 f(t,x)$ (given).

Then by mean value theorem $$|g_n(x)|\le \sup_{t\in [a,b]}|\partial_1 f(t,x)|\le g(x)$$ so DCT applies to $g_n(x)$.

Therefore $$\lim_{t_n\to t} \frac{1}{t_n-t}\int_c^d [f(t_n,x)-f(t,x)]\mathrm dx$$ exists and is equal to $$ \int_c^d \lim_{t_n\to t}\frac{f(t_n,x)-f(t,x)}{t_n-t}\mathrm dx=\int_c^d \partial_1f(t,x) \, \mathrm dx$$

I have just reproduced here what I once learnt from Folland's Real Analysis book.

  • $\begingroup$ O.k., now I understand what you are saying. And I agree that the additional assumption of $\partial_1f(t,x)$ being uniformly bounded in $t$ by some $g(x) \in L^1[c,d]$ would be sufficient. Nevertheless, what I have written is true - as I stated - for almost all $t$. If you could provide a counterexample to prove: "Just ∂1f(t,x) being integrable over [c,d] for every fixed t∈[a,b] isn't sufficient." then that would be great! $\endgroup$
    – Sam
    Dec 28 '11 at 10:19
  • $\begingroup$ How can one write $\frac{\mathrm d}{\mathrm dt}\int_c^d f(t,x) \, \mathrm dx= \lim_{t'\to t} \frac{1}{t'-t}\int_c^d f(t',x)-f(t,x) \, \mathrm dx$ before knowing whether the limit exists. $\endgroup$
    – Ashok
    Dec 28 '11 at 10:36
  • $\begingroup$ It's existence is justified for almost all $t$ by the next line after that and the fact that $$t\to \int_c^d \partial_1f(t,x) \, dx$$ is an $L^1$ function (so almost every $t$ is a Lebesgue-point ). $\endgroup$
    – Sam
    Dec 28 '11 at 11:27
  • $\begingroup$ ah I see. Good point. $\endgroup$
    – Ashok
    Dec 28 '11 at 11:34
  • $\begingroup$ If we only have that the derivative is integrable, do we have an a.e t result? attainable only as a weak derivative? $\endgroup$ Sep 6 '16 at 5:06

I have found a non-trivial counterexample:

Let $I = [0,1]$. I will construct a counterexample $f \in L^1(I\times I)$. Showing that additional assumptions along the lines of

  • For every $t_0\in I$ there exists $g\in L^1(I)$ and a neighbourhood $U$ of $t_0$, s.t. for all $t\in U$ we have: $$|\partial_1f(t,x)| \le g(x) \qquad\text{almost everywhere}$$

cannot be dispensed with (if we want the claimed equality to hold for all $t$ rather than almost all $t$).

We define $f: I\times I\to \mathbb R_{\ge 0}$ by

$$f(t,x) = \sum_{n=1}^\infty \chi_{I_n}(x) \beta_n(t)$$


  • $I_n = (2^{-n}, 2^{-n+1})$ and $\chi_{I_n}$ denotes the characteristic function of $I_n$,
  • $\beta_n:I \to \mathbb R$ is differentiable with
    • $0\le \beta_n(t) \le \dfrac{2^n}{n(n+1)}$, $0\le \beta_n'(t)$ for all $t\in I$,
    • $\beta_n(t) = \begin{cases} 0 & 0\le t \le \dfrac{1}{2n} \\ \dfrac{2^n}{n(n+1)} & \dfrac 1n \le t \le 1 \end{cases}$

i.e. every $\beta_n$ is a monotonic differentiable function which is constant near $t=0$.

For a fixed $x$ at most one summand in the definition of $f(t,x)$ is not equal to zero, so there are no problems with convergence.

Thus we obtain $\partial_1f(t,x) = \sum_{n=1}^\infty \chi_{I_n}(x) \beta_n'(t)$ and - using nonnegativity of all summands

$$ \begin{align} \int_0^1\int_0^1 |\partial_1f(t,x)|\, dt\, dx &= \int_0^1\int_0^1 \left(\sum_{n = 1}^\infty \chi_{I_n}(x)\beta_n'(t)\right) \, dt\, dx\\ &= \sum_{n=1}^\infty |I_n| (\beta_n(1) - \beta_n(0)) \\ &= \sum_{n=1}^\infty 2^{-n} \frac{2^n}{n(n+1)} \\ &= \sum_{n=1}^\infty \left(\frac1n - \frac1{n+1}\right) \\ &= 1 < \infty \end{align} $$ Similarly a quick calculation shows $$ \int_0^1\int_0^1 |f(t,x)|\, dt\, dx \le 1 < \infty $$ Therefore $f, \partial_1f \in L^1(I\times I)$. But now we observe $$ \begin{align} \int_0^1 \frac{f(1/m, x)-f(0,x)}{1/m} \, dx &= m \int_0^1 \left(\sum_{n=1}^\infty \chi_{I_n}(x)[\beta_n(1/m) - \beta_n(0)]\right)\, dx \\ &= m\sum_{n=1}^\infty |I_n|\beta_n(1/m) \\ &\ge m \sum_{n=m}^\infty 2^{-n} \frac{2^n}{n(n+1)} \\ &= m\cdot \frac{1}{m} = 1 \end{align} $$ for all $m\in \mathbb N$. In particular, with $\partial_1f(0,x) = \sum_n \chi_{I_n}(x) \underbrace{\beta_n'(0)}_{=0\; \forall n} = 0$ we see:

$$\limsup_{t\to 0} \int_0^1 \frac{f(t, x)-f(0,x)}{t} \, dx \ge 1 \ne 0 = \int_0^1 \partial_1f(0,x) \, dx$$

Showing that $f$ is a counterexample as claimed.


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