Prove that the function $f(x) = \frac{1}{x^p}$ belongs to $L_{1}(1,\infty)$ (Where $L_1$ is the space of functions that are Lebesgue integrable) if and only if $p > 1$.
proof $\Rightarrow$ Suppose that $f(x) = \frac{1}{x^p} \in L^1(1,\infty)$. As I know, if a function $f \in L^1(E)$, it means that $\int_{E} f \, dm < \infty$. But in this case, if $p > 1$, then $\int_{E} f \, dm = \infty$ as $m(1,\infty) = \infty$,Since $(1, \infty)$ is an interval, and the measure of an interval is equal to its length
And for the converse, I have the same problem; I don't know how to do it.
The fact that $m(1,\infty)=+\infty$ does not imply that $f$ is not integrable in $(1,\infty)$, that would mean, for instance, that $L^1(1,\infty)=L^1(\mathbb{R})=\emptyset$, which clearly is not true.
To prove this statement, you just need to remember the definition of improper integral (and that a non-negative function is Lebesgue Integrable in $(1,+\infty)$ iff $\displaystyle\int_{1}^{\infty} f(x) dx$ exists in the Riemann sense).
For $p>1$ we have a primitive of $\dfrac{1}{x^p}$ is $\dfrac{1}{(1-p)x^{p-1}}$ and thus $$\displaystyle\int_{1}^\infty \dfrac{1}{x^p} dx=\lim_{c\to\infty} \displaystyle\int_{1}^c \dfrac{1}{x^p} dx=\lim_{c\to\infty}\dfrac{1}{(1-p)c^{p-1}}-\dfrac{1}{1-p}=\dfrac{1}{p-1}<\infty$$ as $c^{p-1}\to \infty$ being $p>1$, so $\dfrac{1}{x^p}\in L^1(1,\infty)$.
For $p=1$ a primitive of $\dfrac{1}{x}$ is $\ln(x)$, so $$\displaystyle\int_{1}^\infty \dfrac{1}{x} dx=\lim_{c\to\infty} \displaystyle\int_{1}^c \dfrac{1}{x} dx=\lim_{c\to\infty} \ln(c)-\ln(1)=\infty$$
and $\dfrac{1}{x}\notin L^1(1,\infty)$.
If $p<1$ then $x^p<x$ for $x>1$ and consequently, $\dfrac{1}{x^p}>\dfrac{1}{x}$.
As $\displaystyle\int_{1}^\infty \dfrac{1}{x} dx$ diverges, the direct comparison test tells us $\displaystyle\int_{1}^\infty \dfrac{1}{x^p} dx$ does too for $p<1$, that is, $\dfrac{1}{x^p}\notin L^1(1,\infty)$ for $p<1$.