Differentiation under the integral sign for Lebesgue integrable derivative

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.

Edit:

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.

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

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

where

• $I_n = (2^{-n}, 2^{-n+1})$ 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.