Hello everyone here is an exercice left without correction that I'm struggling with .
$(E,\mathbb{A},\mu)$ is a measurable space
We define :
$\varphi : E \rightarrow \mathbb{R}_+ $
as a positive measurable function
and the application $\nu$ definite over $\mathbb{A}$ by :
$$\forall A \in \mathbb{A} \;|\; \nu(A) = \int \varphi 1_Ad\mu$$
( some trouble with latex and the indicator function )
I proved that $\nu $ is a measure on $(E,\mathbb{A})$
1) Given h a non-negative measurable fuction defined on $E$ prove that :
$$\int hd\nu = \int h\varphi d\mu$$
And that
$h$ is integrable over $\nu$ if and only if $\,h\varphi$ is integrable over $\mu$
2) Now we suppose that $(E,\mathbb{A},\mu)= (\mathbb{R},B(\mathbb{R}),\lambda )\;\; $(Borel set) and we define :
$\varphi (x) = e^{-x}1_{[0,+\infty[}(x)$
Here too I proved that $\nu$ is a probability measure
We introduce : $\,h_1(x)=x \; \; $ and $h_2(x)=x^{2}$
And now we have to calculate :
$$\int h_1d\nu \;\;\;and\;\;\; \int h_2d\nu$$
3) Now we introduce : $\, h_t(x)=e^{tx}$
I found that for $t$<$1\;$ $h_t$ is Lebesgue integrable , but stuck to calculate :
$$\int h_td\nu$$
I hope everything is clear thanks in advance for your help .
Edit : I'm new here is anything wrong with my question ?
You have that $$ \begin{align*} \int h_t\,\mathrm d \nu &=\int h_t\varphi \,\mathrm d \lambda \\ &=\int\mathbf{1}_{[0,\infty )}(x) e^{tx}e^{-x}\,\mathrm d \lambda (x)\\ &=\int_{[0,\infty )} e^{(t-1)x}\,\mathrm d \lambda (x)\\ &\overset{(*)}{=}\int_0^\infty e^{(t-1)x}\,\mathrm d x\\ &=\frac1{t-1}(\lim_{x\to \infty }e^{(t-1)x}-e^{(t-1)0})\\ &=\frac1{t-1}(\lim_{x\to \infty }e^{(t-1)x}-1) \end{align*} $$ where in $(*)$ we used the equivalence of the integral w.r.t. the Lebesgue measure of Riemann-integrable non-negative functions and the improper integral of Riemann, what you can easily prove that it holds using, by example, the monotone convergence theorem.
Thus for $t< 1$ we have that $\int h_t\,\mathrm d \nu =\frac1{1-t}$.