Is this version of the dominated convergence theorem true?

Question

Let $I$ be a real interval, $t_0 \in I,$ $E$ a measurable set in $\mathbb{R}$ and $f \colon I \times E \to \mathbb{R}$ a function such that:

1. $f(t, \cdot) \in L^1(E)$ for every $t \in I;$
2. There exists $\lim_{t \to t_0} f(x, t) =: F(x) \in \mathbb{R}$ for almost every $x \in E;$
3. $F \in L^1(E).$

Can we conclude that $$\lim_{t \to t_0} \int_E f(t, x) \, dx = \int_E F(x) \, dx?$$

It seems to me that it should be true, but I don't see how it can be a consequence of the dominated convergence theorem.

Answer 1

Let $I=E=[0,1]$ and $f(t,x)=\frac 1 t x^{1/t}$ for $t >0$ and $f(0,x)=0$. Take $t_0=0$ Then $F(x)=0$ which is integrable. However a simple calculation shows that $\int f(t,x) dx \to 1$ and $t \to t_0$.

Answer 2

The answer is no: take any $L^1$ function with compact support and integral $1$, say $g(x)$. Then, $f(t,x) = g(x-t)$ is in $L^1$ for all $t$, and converges pointwise to $0$, which is in $L^1$. But for all $t$, $\int f(x,t)\mathrm{d}x = 1$.