Bounding Problem /Conditions for the Lebsgue Integral of a Function depending on two parameters to be continuous

Question

let $F(t)=\int_{E} f_t(x)$ for $t\in J \subseteq \mathbb{R}$.

Then some theorem says that $F$ is continuous if

$1)$ $\forall t_0$ $f_t(x) \rightarrow f_{t_0}(x)$ as $t \rightarrow t_0$ almost everywhere on $E$

or in other words $f$ is continuous with respect to $t$ I think.

$2)$ $\exists$ $g(x) \in L^1(E): |f_t(x)|<g(x)$

so there must be at least one $g$ which works for every $t$

I get this theorem and I get the proof also but when it comes to applying it for some strange reason I can never find a $g$...

e.g for $E=(0,\infty)$ $f_t(x)=x^{t-1}e^{-x^2}$

The best I could do was say it was less than $x^{t-1}$

For this it is $t<1$ which makes the power of $x$ negative that causes problems in $(0,1)$ since it explodes in that region. $t>1$ then becomes a problem in $[1,\infty)$

or

$f_t(x)=\displaystyle\frac{e^{-x}\sin(tx)}{x\sqrt{x}}$

Rewrote as $\displaystyle\frac{te^{-x}\sin(tx)}{(tx)\sqrt{x}}$ But $\frac{\sin(xt)}{(xt)}<1$ so we just need to bound $\displaystyle\frac{te^{-x}}{\sqrt{x}}$ which obviously explodes near $0$. Surely a bound doesn't doesn't exist for either of these? For the first case independence of $t$ is troublesome. and for the second case just straight out dont know how to bound it even with the usual trick of considering cases $(0,1)$ and $[1,\infty)$

Answer 1

First of all, you don't need a continuous $g$, but only $g∈ L^1$. This allows you to piecewise define your dominating function.

Secondly, in the first function the range of $t$ you are allowed to use is $t>0$, roughly speaking because $1/x$ is not integrable around 0. To prove that $\Gamma : (0,∞) → \Bbb R$ is continuous, it suffices to show that the restriction $\Gamma : [ε,∞) → \Bbb R$ is, for every $\varepsilon$. You may use $g(x) := \begin{cases} x^{ε - 1} & x\leq 1 \\ C e^{-x/2} & x > 1 \end{cases}$ where $C<∞ $ is the maximum of $ \tilde g(x) := x^{ε - 1}e^{-x/2}$ on the region $[1,∞)$; it is easy to check that $g∈ L^1$ and also $f_t(x) \leq g(x)$, for $t∈ [\varepsilon,∞)$.

You would have solved the second question by fleshing out the details in the above; near $0$, (a constant multiple of) $1/\sqrt{x}$ is a good choice of dominating function despite being unbounded because it is integrable (with anti-derivative $2\sqrt{x}$)

Let me quickly verify that $g∈ L^1$ for the first function, and you can adapt this for the second function. $$∫ |g| = ∫ g = ∫_0^1 x^{ε -1} \ dx + ∫_1^∞ C e^{-x/2} \ dx $$ the first integral is by Monotone Convergence, $\lim_{N→\infty} ∫_{1/N}^1 x^{ε - 1} \ dx =\lim_{N→∞} \frac{1}{ε}(1 - \frac{1}{N^ε}) = \frac{1}{ε}$ (where we crucially used that $ε>0$). The second integral by high-school calculus is $2Ce^{-1/2}$. $C$ is finite because $\lim_{x→+∞} x^{ε - 1}e^{x/2} = 0$.