Doubts regarding $L_p$ space I had to dive into the basic definitions and properties of $L_p$ spaces as part of a course project I am doing. (More specifically while I was trying to understand Barbalet's Lemma). It was my first time in this topic and I have got really confused now. It would be great if you could answer some of my questions with a numerical example if possible.


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*What exactly is an essentially bounded function? 
The definition I know of is:-  
function $f:[0,1]→ℝ $ is called essentially bounded if there is a number $M $ such that $|f(x)|≤M $ for almost all $ x∈(0,1)$. (That is, the inequality holds on some set $E$ such that $(0,1)$∖E has zero measure.)
Why is $f(x)=x^{−1/(p+1)}$ which defines an element of $L_p((0,1))$ not essentially bounded.
I would also like to ask what is the use of equivalence class of function in such cases?

*Also, what would be the set consisting of "almost everywhere" in any appropriate example (maybe above one)?

*Does essentially bounded function imply it belongs to $L_{\infty}$ space?

*How can a function have $\|f\|_p = 0$ but $f \ne 0$. 
Please explain as if I don't know anything. I am not very sure how much I have understood. I am asking the question only after reading many answers and online lecture notes.
 A: 1) An essentially bounded function is exactly what you described. The $f$ you gave is not essentially bounded, because for any real number $M$, there exists $\varepsilon>0$ such that $f(x)>M$ for all $x\in (0,\varepsilon)$. This shows that it is not essentially bounded because $(0,\varepsilon)$ has positive measure.
2) You wrote in part (1) that "the inequality holds on some set $E$ such that $(0,1)\setminus E$ has zero measure". This means precisely that the inequality holds almost everywhere. Your example doesn't apply because that $f$ is not essentially bounded. Consider the function $g:(0,1)\to\mathbb{R}$ given by
$$
g(x) = \begin{cases}
n & \text{if}\ x=1/n\ \text{for some}\ n\in\mathbb{N},\,n\geq 2 \\
0 & \text{otherwise}.
\end{cases}
$$
Then $g(x)=0\leq 1$ for all $x\in(0,1)\setminus\{1/n\mid n\in\mathbb{N},\,n\geq 2\}$ and $\{1/n \mid n\in\mathbb{N},\,n\geq2\}$ is a measure zero set. Thus $g=0$ a.e., which is another way of saying that the equivalence class of $g$ is equal to the equivalence class of the zero function $0$ under the "equal a.e." equivalence relation.
3) Yes, by definition $L_\infty(0,1)$ is the space of all essentially bounded functions from $(0,1)$ into $\mathbb{R}$. Some authors consider the quotient of this space by the "equal a.e." equivalence relation, but in practice the two are the same.
4) Do you mean $\|f\|_p=0$ but $f\ne0$? Consider the function $g$ given in (2). Then $\|g\|_p=0$, but $g\ne0$. Now we do have $g=0$ a.e., so some would write $g=0$ and it would be understood that the equality here is with respect to the "equal a.e." equivalence relation.
