We consider here a measure space $(\Omega, \mathcal A, \mu)$ and $T \subset \mathbb R$ a topological subspace. For a map $f : T \times \Omega \to \mathbb R$ such that for all $t \in T$ the map $$\begin{array}{l|rcl} f(t, \cdot) : & \Omega & \longrightarrow & \mathbb R \\ & \omega & \longmapsto & f(t,\omega) \end{array}$$ is integrable, one can define the function $$\begin{array}{l|rcl} F : & T & \longrightarrow & \mathbb R \\ & t & \longmapsto & \int_\Omega f(t,\omega) \ d\mu(\omega) \end{array}$$

Following theorem is well known (and can be proven using dominated convergence theorem):

THEOREM for an adherent point $x \in T$, if

• $\forall \omega \in \Omega \lim\limits_{t \to x} f(t,\omega) = \varphi(\omega)$
• There exists a map $g : \Omega \to \mathbb R$ such that $\forall t \in T, \, \forall \omega \in \Omega, \ \vert f(t,\omega) \vert \le g(\omega)$

then $\varphi$ is integrable and $$\lim\limits_{t \to x} F(t) = \int_\Omega \varphi(\omega) \ d\mu(\omega)$$
In other words, one can switch $\lim$ and $\int$ signs.

We provide here a counterexample showing that the conclusion of the theorem might not hold if $f$ is not bounded by a function $g$ as supposed in the premises of the theorem.

Let’s consider $$\begin{array}{l|rcl} f : & [0, \infty) \times [0, \infty) & \longrightarrow & \mathbb R \\ & (t, \omega) & \longmapsto & t e^{-t \omega} \end{array}$$ For all $t \in [0,\infty)$, the map $\omega \mapsto f(t,\omega)$ is integrable on $[0, \infty)$. Therefore, one can define $$\begin{array}{l|rcl} F : & [0, \infty) & \longrightarrow & \mathbb R \\ & t & \longmapsto & \int_0^\infty t e^{-t \omega} \ d\omega\end{array}$$

Moreover for all $\omega \in [0,\infty)$, $f(0,\omega)=0$ hence $F(0)=0$. And for $t \neq 0$, $$F(t)=\int_0^\infty t e^{-t \omega} \ d\omega = 1.$$

Finally $$\lim\limits_{t \to 0} \int_0^\infty t e^{-t \omega} \ d\omega = 1 \neq 0 = \int_0^\infty \lim\limits_{t \to 0} \left( t e^{-t \omega} \right) \ d\omega$$

One can verify that for $\omega \in (0,\infty)$, the maximum of $t \mapsto t e^{-t \omega}$ is obtained for $t = \frac{1}{\omega}$ and is equal to $\frac{1}{e \omega}$. Hence a map $g$ as defined in the hypotheses of the theorem is forced to satisfy the inequality $\frac{1}{e \omega} \le g(\omega)$ and cannot be integrable.