I am trying to understand the following homework.
Let $s>0$ and $u: \Bbb{R}\rightarrow \Bbb{R}$ be: $$ u(x)= \begin{cases} 1/x^s, & 0< x\leq 1 \\ 0,& \text{otherwise.} \end{cases} $$ I need to show that
$u\in \mathcal{M}_\mathbb{R}^+(\mathcal{B}(\mathbb{R}))$ for all $s>0$
and determine for which $s>0$ that $u\in \mathcal{L}^1(\lambda)$
For (1) My idea is that it is a continuous function in the interval and therefore it is Borel measurable and furthermore always positive.
For (2) I think that $u \in \mathcal{L}^1(\lambda)$ when $s\geq 2$ because it is then a convergent series. But my problem is that $u\rightarrow\infty$ when $x\rightarrow 0$. I know there exists unbounded function that are Lebesgue integrable but I don't know how to determine or show it for this function.
Any hints would be appreciated
Your idea about the measurability is correct.
Now i pressume that you mean with $s>0$ that $u(x)=\frac{1}{x^s}$ when $x \in (0,1]$
For $0<s<1$ the function is integrable and you can also compute its integral.
For $s \geq 1$ we have that $$\int_0^1 u(x)dx \geq \sum_{k=1}^{\infty}\int_{\frac{1}{2n+2}}^{\frac{1}{2n+1}}\frac{1}{x^s}dx$$ $$ \geq \sum_{k=1}^{\infty} (2n+1)^s\frac{1}{(2n+1)(2n+2)}$$ $$= \sum_{k=1}^{\infty}\frac{(2n+1)^{s-1}}{2n+2}$$ $$\geq \sum_{k=1}^{\infty} \frac{(2n)^{s-1}}{4n}=+\infty$$ since $s-1 \geq 0$