I'm having trouble parsing the meaning of the following question and am curious if somebody could confirm that the way I am interpreting the prompt is correct:
Let $f(x,t)$ be a real valued function on $\mathbb{R}^2$ such that for each fixed $x$ it is continuous in $t$ and for each $t$ it is Lebesgue integrable in $x$. In addition suppose that $|f(x,t)| \le g(x)$ for $t$ in some open interval $(a,b)$ and some $g$ integrable in $x$. Prove that the function $F(t) = \int_R f(x,t)dx$ is continuous at any point $t_0 \in (a,b)$.
(i) "Let $f(x,t)$ be a real valued function on $\mathbb{R}^2$ such that for each fixed $x$ it is continuous in $t$"
Here I take it that $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ whereby if we fix $x$ and let $p = (x,y)$ that $\forall \epsilon > 0, \exists \delta > 0$ s.t. for any $q = (x,y_1)$ we have $0 < |p - q| < \delta \implies |f(p) - f(q)| < \epsilon$.
(ii) "and for each $t$ it is Lebesgue integrable in $x$"
Here I assume this means that if we fix $t$ and consider $E = \{(x,t) : x \in \mathbb{R} \}$ that $-\infty < \int_E f < +\infty$.
(iii) "In addition suppose that $|f(x,t)| \le g(x)$ for $t$ in some open interval $(a,b)$ and some $g$ integrable in $x$."
Here I assume this means that we fix $(a,b)$ and consider that $\forall t \in (a,b)$, we have that $\forall x \in \mathbb{R}$, $|f(x,t)| \le g(x)$ (implying immediately that $0 \le g$) with $\int g < + \infty$.
(iv) Prove that the function $F(t) = \int_R f(x,t)dx$ is continuous at any point $t_0 \in (a,b)$.
Here I assume that $R \ne \mathbb{R}$ but rather $R = \{(x,t) : x \in \mathbb{R}$ for fixed $t \in \mathbb{R} \}$. Then we aim to show that if $t_0 \in (a,b)$, then $\forall \epsilon > 0$, $\exists \delta > 0$ s.t. $\forall r \in \mathbb{R}$ we have
$0 < |t_0 - r| < \delta \implies |F(t_0) - F(r)| < \epsilon$
or equivalently if $\{t_n\}$ is a sequence of reals which converges to $t_0$, we have that
$\lim_{t_n\to t_0} F(t_n) = F(t_0)\text{ or }\lim_{t_n\to t_0} \int f(x,t_n)dx = \int f(x,t_0)dx$