Continuity of Lebesgue Integral of a Function in 2 Variables

Question

If $(X,M,\mu)$ is a measure space and $f: X \times [a,b] \to \mathbb{C}$ is s.t $\forall t\in[a,b]$ $f(\cdot,t) \in L^1(d\mu)$ ($\int_X|f(x,t)|d\mu(x) < \infty$)

Let $F: [a,b] \to \mathbb{R}$ be defined as $F(t) = \int_X f(x,t)d\mu(x)$.

Suppose that for some $g \in L^1(d\mu)$ $\forall x,t |f(x,t)| < g(x)$ and that $\forall x\in X$ $lim_{t \to t_0}f(x,t) = f(x,t_0)$ then show $lim_{t \to t_0}F(t) = F(t_0)$.

My attempt:

My initial idea was to try to prove this for sequences of rationals in $[a,b]$ and try to generalize:

Let $\{t_n\} \in [a,b] \cap \mathbb{Q} $ converge to $t_0$

If we define $f_n(x,t_n) = f(x,t_n)$ then $f_n : X \to \mathbb{C}$ and $f_n$ converges point wise in $x$ to $f(x,t)$ by the assumption.

Then by the dominated convergence theorem:

$lim_{t\to t_0} \int_Xf(x,t) = lim_{n\to \infty} \int_X f_n(x,t_n)d\mu(x) = \int_Xf(x,t)$

is this argument ok? and is it fair to say that proving this for rational sequences is enough as every irrational is the limit of a sequence of rationals?

Answer 1

Your basic idea is correct: this is essentially a dominated convergence argument. Here are my two corrections:

1. The definition $f_n(x,t_n) = f(x,t_n)$ doesn't make much sense. Sure, it's a reasonable formula, but in this context $f_n$ appears to be a function of two variables and therefore should have a domain like $X\times\mathbb{R}$, which leaves the function ill-defined for values of $t$ other than $t_n$. In any case, you want $f_n$ to have domain $X$ so you can compare to $g\in L^1(X)$. The correct thing to do is set $f_n(x) = f(x,t_n)$.

2. Your dominated convergence argument uses nothing special about the rational numbers to prove convergence of the integral. This suggests that you can take $t_n$ to be any arbitrary sequence of real numbers converging to $0$. This in fact is true: you don't need any weird tricks with rational numbers for this proof.

Answer 2

You're on the right track and you've done all of the heavy lifting. We know the result is true for every sequence and we need show convergence more generally. Approximating by rationals doesn't quite work as far as I can tell so try the following instead: Suppose the contrary so that $F(t)\not\rightarrow F(t_0)$ so that there exists some $\epsilon>0$ such for every $\delta>0$ there exists $t$ with $|t-t_0|<\delta$ and $|F(t)-F(t_0)|>0$. Then setting $\delta=1/n$ we get a sequence $t_n\not\rightarrow t_0$ such that $F(t_n)\not\rightarrow F(t)$ which contradicts you're work above.