Formality of a statement in a calculus exercise, proof check Let $f:\mathbb{R}\rightarrow \mathbb{R} $ be a continuous function positive in the interval $[0,1]$ such that $\exists x_0\in[0,1]$ for which $f(x_0)\gt 1$. Prove that: $$\lim_{n\to \infty}\int_{0}^{1}{f^n(x)dx=+\infty}$$
My work:
$\exists x_0\in[0,1]$ for which $f(x_0)\gt 1$ $\implies$ $\exists I_{x_0}=(x_0-\delta,x_0+\delta)\subseteq[0,1]$ for which $f(x)\ge 1+\epsilon \;\;\; \forall x\in I_{x_0} $ and some $\epsilon\gt 0.$
*(in particular, this is the statement which I'd like to know if it's fine or it needs a formal proof)
From the former inequality we also have: $$f^n(x)\ge(1+\epsilon)^n \;\;\forall x\in I_{x_0} $$
Thus: $$\int_{0}^{1}{f^n(x)dx}\ge(1+\epsilon)^n\cdot2\delta$$
Passing to the limit: $$\lim_{n \to \infty}\int_{0}^{1}{f^n(x)dx}\ge\lim_{n\to \infty}(1+\epsilon)^n\cdot2\delta=+\infty
 $$
Which implies that: $$\lim_{n\to \infty}\int_{0}^{1}{f^n(x)dx=+\infty}$$
 A: There is a problem in the part you are concerned about wher eyou say $f(x) \geq 1+\epsilon$ for all $x \in I_{x_0}$. We cannot lower bound $f(x)$ by $1+\epsilon$ for any arbitrary $\epsilon$.
What we know is that for any $\epsilon>0$, there exists $\delta>0$ such that for all $x$ satisfying $|x-x_0|<\delta$, we have $|f(x)-f(x_0)|<\epsilon$. In particular, this means that $f(x_0)-\epsilon<f(x)<f(x_0)+\epsilon$ for all $x \in I_0:=(x_0-\delta,x_0+\delta)$. So we can conclude that $f(x)>f(x_0)-\epsilon \geq 1-\epsilon$. This inequality doesn't look good enough because we need $f(x)$ to be strictly greater than $1$. The trick is we have to choose a particular $\epsilon$ that will give us a good enough inequality.
Recall that $f(x_0)$ is assumed to be strictly greater than $1$. In other words, $f(x_0)-1>0$. So choose $\epsilon:=\frac{f(x_0)-1}{2}>0$. Then for all $x \in I_{x_0}$, we have $f(x)>f(x_0)-\frac{f(x_0)-1}{2} = 1+\frac{f(x_0)-1}{2}>1$.
There is one other small issue, which is when you say $\int_0^1 f^n(x) \ dx \geq (1+\epsilon)^2 \cdot 2\delta$. This assumes that the interval $I_{x_0}$ is entirely contained in $[0,1]$. You can take care of this by just choosing a small enough $\delta$ so that $I_{x_0}\subseteq [0,1]$.
A: If you have two sequences $(a_{n}),(b_{n})$ such that $a_{n}\geq b_{n}$ and you are not sure about the limit (in the extended real sense) of $a_{n}$ exists or not, you should fornally write $\liminf_{n}$ or $\limsup_{n}$:
\begin{align*}
\liminf_{n}\int_{0}^{1}f_{n}(x)dx\geq\liminf_{n}(1+\epsilon)^{n}\cdot 2\delta=\lim_{n}(1+\epsilon)^{n}\cdot 2\delta=\infty.
\end{align*}
So 
\begin{align*}
\limsup_{n}\int_{0}^{1}f_{n}(x)dx\geq\liminf_{n}\int_{0}^{1}f_{n}(x)dx=\infty,
\end{align*}
we have
\begin{align*}
\lim_{n}\int_{0}^{1}f_{n}(x)dx=\infty.
\end{align*}
