Explore Measurement Concepts with Our Examples and Questions

Let $\mathcal{G}=\{G\subseteq <!--\; \subseteq \; -->{\mathbb{R}}_0^{+}\mid <!--\; \mid \; -->G$ denumerable or $\stackrel{¯<!--\; ¯\; -->}{G}$ denumerable $\}$ and $f:{\mathbb{R}}^{+}⟶<!--\; ⟶\; -->(0,1):x\mapsto <!--\; \mapsto \; -->\{\begin{array}{ll}\exp <!--\; \; -->(-<!--\; -\; -->x)& x\text{ rational }\\ 0& \text{ else }\end{array}$
I am wondering if the function $f$ is $\mathcal{G}-<!--\; -\; -->\mathcal{B}((0,1))$ measurable, because as the $exp(-<!--\; -\; -->x)$ is continuous and $\mathbb{Q}$ is denumerable, normally all $f^{-<!--\; -\; -->1}\left(\mathcal{B}\mathcal{(}\mathcal{(}\mathcal{0}\mathcal{,}\mathcal{1}\mathcal{)}\right)\subset <!--\; \subset \; -->\mathcal{G}$ for the exponential function, but I am lacking the final idea to finish the proof. My idea was to use the following: As $\mathcal{B}((0,1))$ is created by $\mathrm{\sigma <!--\; \sigma \; -->}(\mathcal{E})$ with $\mathcal{E}$=(0,1) then we can use $\mathcal{E}$ to proof that
$f^{-<!--\; -\; -->1}(\mathcal{E})\subset <!--\; \subset \; -->\mathcal{G}$
Any help is much appreciated.

I am a graduate student of Mathematics.In the book Measure,Integration and Real Analysis by Sheldon Axler there is a question that asks the reader to show that there exists no measure space $(X,\mathcal{S},\mathrm{\mu <!--\; \mu \; -->})$ such that $\{\mathrm{\mu <!--\; \mu \; -->}(E):E\in <!--\; \in \; -->\mathcal{S}\}=[0,1)$.I am not sure how to do it.I was thinking of taking a nested sequence $\{E_n\}$ of sets in $\mathcal{S}$ such that $\mathrm{\mu <!--\; \mu \; -->}(E_n)=1-<!--\; -\; -->1/n$ but I don't think that will work.Can someone give me any clue?

Problem:
Assume $0<\mathrm{ϵ<!--\; ϵ\; -->}<1$ and ${\mathrm{\mu <!--\; \mu \; -->}}_{\mathcal{L}}$ is the Lebesgue measure. Find a measurable set $A\subset <!--\; \subset \; -->[0,1]$ such that the closure of A is [0,1] and ${\mathrm{\mu <!--\; \mu \; -->}}_{\mathcal{L}}(A)=\mathrm{ϵ<!--\; ϵ\; -->}$.
My Attempt:
I quickly started reading about the fat Cantor set, whose essential feature (as far as I know) is that its measure can be any such $\mathrm{ϵ<!--\; ϵ\; -->}$. But I think the fatal flaw is that the fat Cantor set is closed, whereas I need a set whose closure is [0,1].
I don't know where to look for a solution. Can I fiddle with fat Cantor sets and solve this problem or do I need a fundamentally different approach?

I have to determine the set which satisfies $\{p\in <!--\; \in \; -->[0,\mathrm{\infty <!--\; \infty \; -->}]:f\in <!--\; \in \; -->{\mathcal{L}}^p(\mathrm{\lambda <!--\; \lambda \; -->})\}$ for a function $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ defined by
$f(x)=\{\begin{array}{ll}\frac{1}{x}& \text{if}x\in <!--\; \in \; -->(0,1]\\ 0& \text{if}x\in <!--\; \in \; -->\mathbb{R}\setminus <!--\; \setminus \; -->(0,1]\end{array}$
I have looked at three different cases, i.e. $p=0,p=(0,\mathrm{\infty <!--\; \infty \; -->})$ and $p=\mathrm{\infty <!--\; \infty \; -->}$ where the last case is troubling me.

If $p=\mathrm{\infty <!--\; \infty \; -->}$ we have to check that $f\in <!--\; \in \; -->{\mathcal{L}}^{\mathrm{\infty <!--\; \infty \; -->}}$. So assume that this is the case. Then by definition there $\mathrm{\exists <!--\; \exists \; -->}R>0$ such that $|f|<R$. We have that
$|f|<R\iff <!--\; \iff \; -->\left|\frac{1}{x}\right|=\frac{1}{x}<R\iff <!--\; \iff \; -->\frac{1}{R}<x\iff <!--\; \iff \; -->x\in <!--\; \in \; -->(\frac{1}{R},1]$
but how do I proceed from here?

While studying concepts of measurable functions, there was a theorem suggesting that if any measurable function is altered on a null set, its measurability still remains.
If $f:E\to <!--\; \to \; -->\mathbb{R}$ is measurable, $E\in <!--\; \in \; -->\mathcal{M},g:E\to <!--\; \to \; -->\mathbb{R}$ is such that the set $\{x:f(x)\ne <!--\; \ne \; -->g(x)\}$ is null, then $g$ is measurable.
They showed the following proof.
Consider the difference $d(x)=g(x)-<!--\; -\; -->f(x)$. It is zero except on a null set, so
$\{x:d(x)><!--\; >\; -->a\}=\{\begin{array}{ll}\text{a null set}& a\ge <!--\; \ge \; -->0\\ \text{a full set}& a<<!--\; <\; -->0\end{array}$
Here, a full set is the complement of a null set. Since both null and full sets are measurable, $d$ is a measurable function. $g=f+d$ is thus measurable.
Now, I was curious whether the statement remains true if I change the set $\{x:f(x)\ne <!--\; \ne \; -->g(x)\}$ into $\{x:f(x)=g(x)\}$ , i.e., differ at points in a full set, because there was no doubt if I alter the proof as the following.
The difference $d(x)$ is still measurable since
$\{x:d(x)><!--\; >\; -->a\}=\{\begin{array}{ll}\text{a full set}& a\ge <!--\; \ge \; -->0\\ \text{a full set}& a<<!--\; <\; -->0\end{array}$
However, the later statement is actually not plausible at all. Is there any contradictory logic among here?
Appreicate as always.

I wonder why uncertainties in angle measurement MUST be in radians. For example, I want to calculate the uncertainty in measuring the function $y=\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})$ when the angle is measured $\pm <!--\; \pm \; -->1$ degree. I do this using differential, so $dy=\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})d\mathrm{\theta <!--\; \theta \; -->}$, now $d\mathrm{\theta <!--\; \theta \; -->}=\pm <!--\; \pm \; -->1$ degree is the error in $\mathrm{\theta <!--\; \theta \; -->}$. Now, all the course notes/ books I read says this must be converted in radians, even though the angle we use here is measured in degree. How come?

Let $f:X\to <!--\; \to \; -->\mathbb{R}$ be a continuous function on a compact metric space $X$. Assume that a Borel probability measure $\mathrm{\mu <!--\; \mu \; -->}$ is absolutely continuous with respect to Lebesgue measure Leb.
Is it true that if $f(x)<0$ for Leb a.e. x, then $\int <!--\; \int \; -->fd\mathrm{\mu <!--\; \mu \; -->}<0?$?
I think it should be true as $\mathrm{\mu <!--\; \mu \; -->}<<\text{Leb}$.
Attempt: Assume that $\int <!--\; \int \; -->fd\mathrm{\mu <!--\; \mu \; -->}\ge <!--\; \ge \; -->0$. Then $f\ge <!--\; \ge \; -->0$ μ a.e. x. That means $\mathrm{\mu <!--\; \mu \; -->}(\{x:f(x)>0\})>0$. That implies $\text{Leb}((\{x:f(x)>0\})>0$ which is not true.

Let $u\in <!--\; \in \; -->{\mathcal{L}}^{\mathcal{1}}\mathcal{(}\mathrm{\mu <!--\; \mu \; -->}\mathcal{)}$, $T$ A measure preserving map s.t.: $\int <!--\; \int \; -->(u)d\mathrm{\mu <!--\; \mu \; -->}=\int <!--\; \int \; -->u\circ <!--\; \circ \; -->Td\mathrm{\mu <!--\; \mu \; -->}$. $A_{n,\mathrm{ϵ<!--\; ϵ\; -->}}=\{\frac{|u(T^n(x))|}{n^2}><!--\; >\; -->\mathrm{ϵ<!--\; ϵ\; -->}\}$. $T^n=T\circ <!--\; \circ \; -->T\circ <!--\; \circ \; -->T\dots$ n times.
Prove $\underset{n\ge <!--\; \ge \; -->1}{\sum <!--\; \sum \; -->}\mathrm{\mu <!--\; \mu \; -->}(A_{n,\mathrm{ϵ<!--\; ϵ\; -->}})$ is finite.
My attempt:
By Markov inequality we get: $\mathrm{\mu <!--\; \mu \; -->}(A_{n,\mathrm{ϵ<!--\; ϵ\; -->}})\le <!--\; \le \; -->\frac{1}{\mathrm{ϵ<!--\; ϵ\; -->}}\int <!--\; \int \; -->\frac{|u(T^n(x))|}{n^2}d\mathrm{\mu <!--\; \mu \; -->}$, since $T$ is measure preserving we get: $|u(T^n(x))|\in <!--\; \in \; -->{\mathcal{L}}^{\mathcal{1}}\mathcal{(}\mathrm{\mu <!--\; \mu \; -->}\mathcal{)}$. Summing over n gives:
$\underset{n\ge <!--\; \ge \; -->1}{\sum <!--\; \sum \; -->}\mathrm{\mu <!--\; \mu \; -->}(A_{n,\mathrm{ϵ<!--\; ϵ\; -->})}\le <!--\; \le \; -->\underset{n\ge <!--\; \ge \; -->1}{\sum <!--\; \sum \; -->}{\mathrm{ϵ<!--\; ϵ\; -->}}^{-<!--\; -\; -->1}{n}^{-<!--\; -\; -->2}\int <!--\; \int \; -->|u(T^n(x))|d\mathrm{\mu <!--\; \mu \; -->}.$
This is the part where I am stuck. I know that all the integrals are finite but how do I ensure that summing over all the integrals stays finite? Can I make the integral independent of n? Or can I just state by assumption: $\int <!--\; \int \; -->|u(T^n(x))|d\mathrm{\mu <!--\; \mu \; -->}=\int <!--\; \int \; -->|u|d\mathrm{\mu <!--\; \mu \; -->}$?

To calculate the random error in a set of measurements this is what I would do.

1. Get the standard deviation of the measurements:
$\mathrm{\sigma <!--\; \sigma \; -->}=\sqrt{\frac{1}{N-<!--\; -\; -->1}{\mathrm{\Sigma <!--\; \Sigma \; -->}}_{i=1}^N(x_i-<!--\; -\; -->\stackrel{¯<!--\; ¯\; -->}{x}{)}^2}$
Where $N$ is the number of repeats, $x_i$ is the value of each sample, $\stackrel{¯<!--\; ¯\; -->}{x}$ is the mean, and $\mathrm{\sigma <!--\; \sigma \; -->}$ is the standard deviation.
2. The standard error of the mean:
$s_{\stackrel{¯<!--\; ¯\; -->}{x}}=\frac{\mathrm{\sigma <!--\; \sigma \; -->}}{\sqrt{N}}$
3. Obtain the random error using a t distribution:
${\mathrm{\epsilon <!--\; \epsilon \; -->}}_{rnd}=\pm t_{N-<!--\; -\; -->1}\ast <!--\; \ast \; -->s_{\stackrel{¯<!--\; ¯\; -->}{x}}$

The problem is, if I have a measurement with a single reading of a given value, this method for random error calculation is clearly no longer applicable.
Therefore, how would you go about calculating the random error in a for a sample set where N=1? Is that even possible?

Let $E$ be a metric space, $(X_t{)}_{t\ge <!--\; \ge \; -->0}$ be an $E$-valued right-continuous process and $f:E\to <!--\; \to \; -->[0,\mathrm{\infty <!--\; \infty \; -->})$ be locally bounded and Borel measurable. Is this enough to ensure that
$\begin{array}{}\text{(1)}& {\int <!--\; \int \; -->}_0^{t}f(X_s)\mathrm{d}s<\mathrm{\infty <!--\; \infty \; -->}\end{array}$
for all $t\ge <!--\; \ge \; -->0$? The question is clearly trivial, when $(X_t{)}_{t\ge <!--\; \ge \; -->0}$ and $f$ are continuous.

Setting We work on a filtered probability space $(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{F},(\mathcal{F}{)}_{t\in <!--\; \in \; -->[0,T]},P)$. Let $(X^n{)}_n$ and X be finite variation processes such that $X^n$ converges pointwise to X, i.e.
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{X}_t^n(\mathrm{\omega <!--\; \omega \; -->})=X_t(\mathrm{\omega <!--\; \omega \; -->})$
for all $(\mathrm{\omega <!--\; \omega \; -->},t)\in <!--\; \in \; -->\mathrm{\Omega <!--\; \Omega \; -->}\times <!--\; \times \; -->[0,T]$, and such that $(\Vert <!--\; \Vert \; -->X^n\Vert <!--\; \Vert \; -->{)}^n$ converges pointwise to $\Vert <!--\; \Vert \; -->X\Vert <!--\; \Vert \; -->$ (where $\Vert <!--\; \Vert \; -->\cdot <!--\; \cdot \; -->{\Vert <!--\; \Vert \; -->}_t$ denotes the total variation on [0,t]).
Question Does this imply that $\underset{n}{sup}\Vert <!--\; \Vert \; -->X^n\Vert <!--\; \Vert \; --><\mathrm{\infty <!--\; \infty \; -->}$ pointwise?

I`m trying to prove that:
$\mathrm{\mu <!--\; \mu \; -->}(A)=\underset{n\in <!--\; \in \; -->\mathbb{N}}{\sum <!--\; \sum \; -->}{\mathrm{\delta <!--\; \delta \; -->}}_{\frac{1}{n}}(A)$
is not a regular measure.
A measure is regular if:
$\mathrm{\mu <!--\; \mu \; -->}(A)=inf\{\mathrm{\mu <!--\; \mu \; -->}(G)|A\subset <!--\; \subset \; -->G,G\text{ is open}\}=sup\{\mathrm{\mu <!--\; \mu \; -->}(F)|F\subset <!--\; \subset \; -->A,F\text{ is closed}\}$
for any measurable $A$.
I'm trying to find a counter example for that but can't come up with any.
I was thinking about a set which contains a finite number of 1/n but can only be covered by $(0,\mathrm{ϵ<!--\; ϵ\; -->})$ for some $\mathrm{ϵ<!--\; ϵ\; -->}>0$, leading to a contradiction.

Suppose that the Hilbert space of a quantum-mechanical system - which we will call the quantum door - is generated by two states, |open> and |closed>, forming an orthonormal basis. Suppose also that the system is prepared in the state
$|\mathrm{\psi <!--\; \psi \; -->}(x)>=\frac{1}{\sqrt{5}}(|OPEN>+2|CLOSED>)$ We are given a device that measures whether the quantum door is open or closed.
(i)If we perform a measurement, which probability do we have to find the quantum door open?
(ii) Suppose the measurement returns that the quantum door is closed, and assume that the quantum Hamiltonian is identically 0 for this system at any future times. Does the door stay closed forever?
For part (i) I get
$P_{Open}=(<open|\frac{1}{\sqrt{5}}(|OPEN>+2|CLOSED>{)}^2$ = 1/5 ?
I also need help with part (ii), i am unure about this.

I'm reading about this in one of the articles that the teacher gave me

Suppose that $(S_1,{\mathrm{\mu <!--\; \mu \; -->}}_1)$ and $(S_2,{\mathrm{\mu <!--\; \mu \; -->}}_2)$ are two $\mathrm{\sigma <!--\; \sigma \; -->}$-finite measure spaces and $F:S_1\times <!--\; \times \; -->S_2\to <!--\; \to \; -->\mathbb{R}$ is measurable. Then
${[{\int <!--\; \int \; -->}_{S_2}{|{\int <!--\; \int \; -->}_{S_1}F(x,y){\mathrm{\mu <!--\; \mu \; -->}}_1(\mathrm{d}x)|}^p{\mathrm{\mu <!--\; \mu \; -->}}_2(\mathrm{d}y)]}^{\frac{1}{p}}\le <!--\; \le \; -->{\int <!--\; \int \; -->}_{S_1}{({\int <!--\; \int \; -->}_{S_2}|F(x,y){|}^p{\mathrm{\mu <!--\; \mu \; -->}}_2(\mathrm{d}y))}^{\frac{1}{p}}{\mathrm{\mu <!--\; \mu \; -->}}_1(\mathrm{d}x)$
with obvious modifications in the case $p=\mathrm{\infty <!--\; \infty \; -->}$. If $p>1$, and both sides are finite, then equality holds only if $|F(x,y)|=\mathrm{\phi <!--\; \phi \; -->}(x)\mathrm{\psi <!--\; \psi \; -->}(y)$ a.e. for some non-negative measurable functions $\mathrm{\phi <!--\; \phi \; -->}$ and $\mathrm{\psi <!--\; \psi \; -->}$.

If $F$ is non-negative, then the function
$f:y\mapsto <!--\; \mapsto \; -->{\int <!--\; \int \; -->}_{S_1}F(x,y){\mathrm{\mu <!--\; \mu \; -->}}_1(\mathrm{d}x)$
and thus $|f{|}^p$ are measurable. For the LHS to be well-defined, the function $|f{|}^p$ should be measurable. In our case, $F$ is not assumed to be non-negative. Could you elaborate on how to prove it?

I'm looking to show that $\frac{1}{|x{|}^n}$ is not integrable on the subset of ${\mathbb{R}}^n$ where $|x|\ge <!--\; \ge \; -->1$. It's easy to do on $\mathbb{R}$, and I think I need to apply Fubini's theorem for the general case, but not sure how to do so in n-dimensional space. Thanks

Evaluate:
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}{(1+\frac{x}{n})}^{-<!--\; -\; -->n}\sin <!--\; \; -->\frac{x}{n}dx.$
I have tried to use the (Lebesgue) dominated convergence theorem to evaluate the same. At first I noticed that:
$\begin{array}{}& |{(1+\frac{x}{n})}^{-<!--\; -\; -->n}\sin <!--\; \; -->\frac{x}{n}|\le <!--\; \le \; -->\left|{(1+\frac{x}{n})}^{-<!--\; -\; -->n}\right|\le <!--\; \le \; -->1\text{ for all positive }x\text{ and for all natural }n.\end{array}$
Since $g(x)=1$ is Lebesgue integrable for each $x\in <!--\; \in \; -->[0,\mathrm{\infty <!--\; \infty \; -->}).$ So by DCT, the given limit equals:
${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{(1+\frac{x}{n})}^{-<!--\; -\; -->n}\sin <!--\; \; -->\frac{x}{n}dx={\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}{e}^{-<!--\; -\; -->x}\cdot <!--\; \cdot \; -->0dx=0.$
This question was asked in our end term exam this semester and so I don't know the correct answer to this question. Therefore, just to check whether I have evaluated the limit correctly I am posting the same here on MSE. If I have gone wrong somewhere, please point out and give some insights. Thanks in advance.

ABC is a triangle in which: P - a = 8 cm , P - b = 6 cm , P - c = 4 cm .
Find measurement of the largest angle in triangle where 2P = a + b+ c
Edit : My answer:
$2P=a+b+c$
$P=\frac{a+b+c}{2}$
$P-<!--\; -\; -->b=\frac{a+b+c}{2}-<!--\; -\; -->b=6$
$=\frac{a+b+c}{2}-<!--\; -\; -->\frac{2b}{2}=6$
$=\frac{a-<!--\; -\; -->b+c}{2}=6$
$=a-<!--\; -\; -->b+c=12$
$P-<!--\; -\; -->a=\frac{a+b+c}{2}-<!--\; -\; -->a=8$
$=\frac{a+b+c}{2}-<!--\; -\; -->\frac{2a}{2}=8$
$=\frac{-<!--\; -\; -->a+b+c}{2}=8$
$=-<!--\; -\; -->a+b+c=16$
$-<!--\; -\; -->a+b+c=16$
$a-<!--\; -\; -->b+c=12$
$2c=28$
$c=14$
$P-<!--\; -\; -->c=4$
$P-<!--\; -\; -->14=4$
$P=18$
$P-<!--\; -\; -->a=8$
$18-<!--\; -\; -->a=8$
$a=10$
$P-<!--\; -\; -->b=6$
$18-<!--\; -\; -->b=6$
$b=12$
$a<b<c$
Then angle C is largest angle
$CosC=\frac{{10}^2+{12}^2-{14}^2}{2•10•12}=\frac{1}{5}$
Then angle
$<C=arccos\frac{1}{5}=78.46$

Let $f_n:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ be defined by $f_n(\mathrm{\omega <!--\; \omega \; -->})=\mathrm{\omega <!--\; \omega \; -->}(1+\frac{1}{n})$ and let $\mathrm{\mu <!--\; \mu \; -->}$ be the Lebesgue measure.
I have to show that $f_n$ does not converge in measure μ to ω.
Here's my attempt: By definition, $f_n$ converges in measure to $\mathrm{\omega <!--\; \omega \; -->}$ if:
${\displaystyle \underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\mathrm{\mu <!--\; \mu \; -->}(\mathrm{\omega <!--\; \omega \; -->}:|f_n(\mathrm{\omega <!--\; \omega \; -->})-<!--\; -\; -->\mathrm{\omega <!--\; \omega \; -->}|\ge <!--\; \ge \; -->\mathrm{\delta <!--\; \delta \; -->})=0.}$
Rewriting, it becomes that:
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\mathrm{\mu <!--\; \mu \; -->}(\mathrm{\omega <!--\; \omega \; -->}:|\mathrm{\omega <!--\; \omega \; -->}|\ge <!--\; \ge \; -->n\cdot <!--\; \cdot \; -->\mathrm{\delta <!--\; \delta \; -->})=0.$
I realized that if $\mathrm{\delta <!--\; \delta \; -->}=1$ for example, then the measure is surely 0 as $n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$. I need to come up with a value of $\mathrm{\delta <!--\; \delta \; -->}$ such that the measure is non-zero. I thought about letting $\mathrm{\delta <!--\; \delta \; -->}=\frac{1}{n},$, but am unsure if it'll become indeterminate once I set $n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$ (because I basically get $\mathrm{\infty <!--\; \infty \; -->}\cdot <!--\; \cdot \; -->\frac{1}{\mathrm{\infty <!--\; \infty \; -->}}).$
Does this solution of $\mathrm{\delta <!--\; \delta \; -->}=\frac{1}{n}$ truly work as a counterexample? If not, what should I do?

Let $f_n\in <!--\; \in \; -->L^1$ be a seq of nonnegative functions s.t $f_n\to <!--\; \to \; -->f$ a.e ptwise. Show $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\int <!--\; \int \; -->(min(f,f_n)d\mathrm{\mu <!--\; \mu \; -->}=\int <!--\; \int \; -->fd\mathrm{\mu <!--\; \mu \; -->}$.
For this, I wanted to use the dominated convergence theorem.
I have shown that $min(f,f_n)$ converges to $f$ pointwise since $|f-<!--\; -\; -->f_n|\ge <!--\; \ge \; -->|f-<!--\; -\; -->min(f,f_n)|$.
Moreover, $|min(f,f_n)|=min(f,f_n)\le <!--\; \le \; -->f$. But I cannot use the dominated convergence theorem directly since the integral of $f$ may not be finite. Is there a way to fix this?

Given:
Let $X$ be a càdlàg Lévy process, $\mathrm{\pi <!--\; \pi \; -->}$ denote the unique random measure with
${\mathrm{\pi <!--\; \pi \; -->}}_{\mathrm{\omega <!--\; \omega \; -->}}([0,t]\times <!--\; \times \; -->B)=\underset{s\in <!--\; \in \; -->[0,t]}{\sum <!--\; \sum \; -->}{1}_B\left(\mathrm{\Delta <!--\; \Delta \; -->}{X}_s(\mathrm{\omega <!--\; \omega \; -->})\right),$
where $\mathrm{\Delta <!--\; \Delta \; -->}x(t):=x(t)-<!--\; -\; -->\underset{s\to <!--\; \to \; -->t-<!--\; -\; -->}{lim}x(s)$,
${\mathrm{\zeta <!--\; \zeta \; -->}}_t(\mathrm{\omega <!--\; \omega \; -->},B):={\mathrm{\pi <!--\; \pi \; -->}}_{\mathrm{\omega <!--\; \omega \; -->}}([0,t]\times <!--\; \times \; -->B)$
and
$\mathrm{\tau <!--\; \tau \; -->}:=inf\{t>0:\mathrm{\Delta <!--\; \Delta \; -->}{X}_t\in <!--\; \in \; -->B\}$
for some Borel measurable $B$ with $0\notin \stackrel{¯<!--\; ¯\; -->}{B}$.

We can show that
$\begin{array}{}\text{(1)}& \mathrm{P}<!--\; \; -->[\mathrm{\tau <!--\; \tau \; -->}>s+t]=\mathrm{P}<!--\; \; -->[\mathrm{\tau <!--\; \tau \; -->}>s]\mathrm{P}<!--\; \; -->[\mathrm{\tau <!--\; \tau \; -->}>t].\end{array}$
Since
$\mathrm{\varphi <!--\; \varphi \; -->}(t):=\mathrm{P}<!--\; \; -->[\mathrm{\tau <!--\; \tau \; -->}>t]$
is right-continuous, this implies that
$\begin{array}{}\text{(2)}& \mathrm{\varphi <!--\; \varphi \; -->}(t)=e^{t\ln <!--\; \; -->\mathrm{\varphi <!--\; \varphi \; -->}(1)}.\end{array}$
By the result above, $\mathrm{\tau <!--\; \tau \; -->}$ is exponentially distributed, but how do we see that $\mathrm{\tau <!--\; \tau \; -->}\sim <!--\; \sim \; -->\mathrm{Exp}<!--\; \; -->(\mathrm{\lambda <!--\; \lambda \; -->})$, where $\mathrm{\lambda <!--\; \lambda \; -->}:=\mathrm{E}<!--\; \; -->[{\mathrm{\zeta <!--\; \zeta \; -->}}_1(\cdot <!--\; \cdot \; -->,B)]$?

Clearly,
$\begin{array}{}\text{(3)}& \mathrm{Exp}<!--\; \; -->(\mathrm{\lambda <!--\; \lambda \; -->})((t,\mathrm{\infty <!--\; \infty \; -->}))=e^{-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}t}.\end{array}$
So, it seems like we need to show
$\begin{array}{}\text{(4)}& \ln <!--\; \; -->\mathrm{\varphi <!--\; \varphi \; -->}(1)=-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}(B).\end{array}$
But how do we do this?