# Measure Theory Question Involving an $f: \mathbb{R}^2 \rightarrow \mathbb{R}$

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

$$\lim_{t_n\to t_0} F(t_n) = F(t_0)\text{ or }\lim_{t_n\to t_0} \int f(x,t_n)dx = \int f(x,t_0)dx$$

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(i) This means that if, for each $x\in\Bbb{R}$, you define $g_{x}(t):\Bbb{R}\rightarrow\Bbb{R}$ by $g_{x}(t)=f(x,t)$, then each $g_{x}(t)$ is continuous.
(ii) As before, consider a fixed $t\in\Bbb{R}$, and define $h_{t}(x)=f(x,t)$. Then this means $\int_{\Bbb{R}}{|h_{t}|\ d\mu}<\infty$, for each $t\in\Bbb{R}$.
(iii) Correct (although your last statement should read $\int{|g|}<\infty$)
(iv) Once again, it's simpler to understand if we go back to writing $h_{t}(x)=f(x,t)$. Then Define $F(t)=\int_{\Bbb{R}}{f(x,t)\ dx}=\int_{\Bbb{R}}{h_{t}(x)\ dx}$ - for each $t$, we give a value to this function by calculating the integral of $h_{t}(x)$ over $\Bbb{R}$. Then this is saying that $F(t)$ is continuous.