$\newcommand{\d}{\,\mathrm{d}}$I here use Royden's definitions of equiintegrability, and tightness:
A family of measurable functions $\{f_n\}_{n\in\Bbb N}:X\to\Bbb R$, where $(X,\mathcal{A},\mu)$ is a measure space, is said to be equiintegrable if: $$\forall\varepsilon\gt0,\,\exists\delta\gt0:\forall A\in\mathcal{A},\,\forall n,\,\mu(A)\lt\delta\implies\int_A|f_n|\d\mu\lt\varepsilon$$
The family is said to be tight if: $$\forall\varepsilon\gt0,\,\exists E\in\mathcal{A},\,\mu(E)\lt\infty:\forall n,\,\int_{X\setminus E}|f_n|\d\mu\lt\varepsilon$$
With that out of the way, and paraphrasing the Wikipedia article to account for my different nomenclature (I use Royden's so as to be able to follow my own notes as I type this), I find that Wikipedia formally requires the family $\{f_n\}$ to be a subset of $\mathcal{L}^p(X)$, yet the proof I have here never requires the $f_n$ to be integrable; for the sake of rigour, I would dearly like to know if it is required that $\{f_n\}\subset\mathcal{L}^p(X)$ for $\lim_{n\to\infty}\|f-f_n\|_{\mathcal{L}^p}=0$.
Vitali's Convergence Theorem, Finite Case (one possible statement – I have seen many slight variants):
Let $\{f_n:n\in\Bbb N\}\subset\mathcal{L}^p(X)$ where $(X,\mathcal{A},\mu)$ is a finite measure space. Suppose $(f_n)$ has a pointwise limit $f$; if $\{f_n\}$ is an equiintegrable family, then $f\in\mathcal{L}^p$ and $\lim_{n\to\infty}\int_Xf_n\d\mu=\int_Xf\d\mu$.
And "my" proof (a combination of Royden's less general work and what I have scrounged from other stack posts):
As this proof immediately generalises if $p\neq1$, w.l.o.g take $p=1$. As $\mu(X)\lt\infty$, Egorov's theorem ensures that $\forall\delta\gt0,\,\exists E\in\mathcal{A}:\mu(E)\lt\delta,\,(f_n)|_{X\setminus E}\rightrightarrows f|_{X\setminus E}$. The $\mu$-finiteness of $X$ also gives that $X\setminus E$ is of finite measure, hence, by uniform convergence: $$\forall\varepsilon\gt0,\,\exists\delta\gt0,\,\int_{X\setminus E}|f-f_n|\d\mu\lt\frac{\varepsilon}{3}$$
Where $E$ is as before and all $n\gt N$ are suitably large, and $\delta$ is chosen suitably small. Fatou's Lemma combined with the equiintegrability of $\{f_n\}$ shows that, if $\delta$ is suitably small: $$\int_E|f|\d\mu=\int_E\liminf_{n\to\infty}|f_n|\d\mu\le\liminf_{n\to\infty}\int_E|f_n|\d\mu\lt\frac{\varepsilon}{3}$$Since equiintegrability requires that if $\mu(E)\lt\delta,\int_E|f_n|\d\mu$ may be made as small as desired, with the same $\delta$ for all $n$.
Taking $\delta$ to be the smaller of the two used above, and for $n$ as large as before, and $E$ as before, then one has:$$\begin{align}\forall\varepsilon\gt0,\,\forall n\gt N:\int_X|f-f_n|\d\mu&\le\int_E|f-f_n|\d\mu+\int_{X\setminus E}|f-f_n|\d\mu\\&\le\int_E|f|\d\mu+\int_E|f_n|\d\mu+\int_{X\setminus E}|f-f_n|\d\mu\\&\lt\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon\end{align}$$And therefore $f_n$ converges to $f$ in $\mathcal{L}^p$ norm.
One deduces that $f\in\mathcal{L}^p(X)$ by a similar argument, that necessitates the integrability of $\{f_n\}$. However, I am fairly sure that one has convergence in $\mathcal{L}^p$ norm anyway, regardless of whether or not $\{f_n\}\subset\mathcal{L}^p$, since Egorov's theorem doesn't require integrability, and I don't believe any of my steps did either.
Using the above definition of tightness, if one requires that $\{f_n\}$ is also tight then the requirement that $\mu(X)\lt\infty$ may be dropped; moreover, in this case, I believe one has that $f$ is $p$-integrable and that $f_n\to f$ in norm regardless of whether or not $f_n$ are themselves $p$-integrable. I will add the proof I have if this statement requires expanding.
The question:
The formal statements require $\{f_n\}$ to be integrable; I believe that to get the convergence-in-norm conclusion, one may drop this requirement, and if one replaces it with $\{f_n\}$ must be tight, one gets all conclusions independent of the integrability of each $f_n$. Is this right?
Best Answer
I think you need to have $\{|f_n|^p\}$ equi-integrable in the statement of the theorem. Using many symbols makes your proof hard to read. The proof can be written without any $\exists$ or $\forall$.
I wrote a proof for $L^p$ myself to check, and I agree that you don't need the hypothesis $f_n \in L^p(X, \mu)$ to deduce that $\lVert f_n - f \rVert_{L^p} \to 0$.
The equiintegrability of $|f_n|^p$ and the finiteness of the measure space do imply that $f_n \in L^p(X, \mu)$, and therefore that $f \in L^p(X, \mu)$.
The tightness hypothesis for $\{|f_n|^p\}$ is just there to reduce to the finite measure case, so the $f_n \in L^p$ hypothesis is redundant here too.