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$\newcommand\R{\mathbb R}$Your claim is incorrect.

E.g., let $\mu$ be the Lebesgue measure over $X:=\R$. Let $$\Phi:=\{f_n\colon\, n\in\mathbb N\},$$ where $$f_n(x):=\frac1{1+(x-n)^2}$$ for real $x$. Then $\Phi$ is uniformly bounded in $L^1$, does not escape to vertical infinity, and does not escape to width infinity; however, $\Phi$ is neither tight nor equi-integrable.

Also, $f_n\to0$ pointwise and hence almost everywhere, but $f_n\not\to0$ in $L^1(\mu)$.

On the the other hand, it is easy to see that, if a subset $\Phi$ of $L^1(\mu)$ does not escape to vertical infinity and is tight, then $\Phi$ is equi-integrable and uniformly bounded in $L^1$. Also, if $\Phi=\{f_n\colon\, n\in\mathbb N\}$ does not escape to vertical infinity and is tight, and if $f_n\to f$ almost everywhere, then $f_n\to f$ in $L^1(\mu)$.

$\newcommand\R{\mathbb R}\newcommand{\ep}{\epsilon}\newcommand{\de}{\delta}$ Your claim is not quite correct.

E.g., let $\mu$ be the Lebesgue measure over $X:=\R$. Let $$\Phi:=\{f_n\colon\, n\in\mathbb N\},$$ where $$f_n(x):=\frac1{1+(x-n)^2}$$ for real $x$. Then $\Phi$ is uniformly bounded in $L^1$, does not escape to vertical infinity, and does not escape to width infinity; however, $\Phi$ is not tight.

Also, $f_n\to0$ pointwise and hence almost everywhere, but $f_n\not\to0$ in $L^1(\mu)$.

On the the other hand, if $\Phi$ does not escape to vertical infinity, then $\Phi$ is equi-integrable (the condition that $\Phi$ does not escape to width infinity is not needed here). Indeed, suppose that $\Phi$ does not escape to vertical infinity, and take any real $\ep>0$. Then there is a real $M>0$ such that $\sup_{f\in\Phi}\int_{|f|\ge M}|f|\,d\mu<\ep/2$. Let now $\de:=\ep/(2M)$. Then for any $f\in\Phi$ and any $A\in\mathcal F$ such that $\mu(A)<\de$ we have \begin{equation} \int_A|f|\,d\mu\le\int_{|f|\ge M}|f|\,d\mu+\int_A M\,d\mu<\ep/2+M\de=\ep. \end{equation} So, $\Phi$ is equi-integrable, as claimed.

Also, if $\Phi=\{f_n\colon\, n\in\mathbb N\}$ does not escape to vertical infinity and is tight, and if $f_n\to f$ almost everywhere, then $f_n\to f$ in $L^1(\mu)$.

$\newcommand\R{\mathbb R}$Your claim is incorrect.

E.g., let $\mu$ be the Lebesgue measure over $X:=\R$. Let $$\Phi:=\{f_n\colon\, t\in\mathbb N\},$$ where $$f_n(x):=\frac1{1+(x-n)^2}$$ for real $x$. Then $\Phi$ is uniformly bounded in $L^1$, does not escape to vertical infinity, and does not escape to width infinity; however, $\Phi$ is neither tight nor equi-integrable.

Also, $f_n\to0$ pointwise and hence almost everywhere, but $f_n\not\to0$ in $L^1(\mu)$.

On the the other hand, it is easy to see that, if a subset $\Phi$ of $L^1(\mu)$ does not escape to vertical infinity and is tight, then $\Phi$ is equi-integrable and uniformly bounded in $L^1$. Also, if $\Phi=\{f_n\colon\, t\in\mathbb N\}$ does not escape to vertical infinity and is tight, and if $f_n\to f$ almost everywhere, then $f_n\to f$ in $L^1(\mu)$.

$\newcommand\R{\mathbb R}$Your claim is incorrect.

E.g., let $\mu$ be the Lebesgue measure over $X:=\R$. Let $$\Phi:=\{f_n\colon\, n\in\mathbb N\},$$ where $$f_n(x):=\frac1{1+(x-n)^2}$$ for real $x$. Then $\Phi$ is uniformly bounded in $L^1$, does not escape to vertical infinity, and does not escape to width infinity; however, $\Phi$ is neither tight nor equi-integrable.

Also, $f_n\to0$ pointwise and hence almost everywhere, but $f_n\not\to0$ in $L^1(\mu)$.

On the the other hand, it is easy to see that, if a subset $\Phi$ of $L^1(\mu)$ does not escape to vertical infinity and is tight, then $\Phi$ is equi-integrable and uniformly bounded in $L^1$. Also, if $\Phi=\{f_n\colon\, n\in\mathbb N\}$ does not escape to vertical infinity and is tight, and if $f_n\to f$ almost everywhere, then $f_n\to f$ in $L^1(\mu)$.

$\newcommand\R{\mathbb R}$Your claim is incorrect.

E.g., let $\mu$ be the Lebesgue measure over $X:=\R$. Let $$\Phi:=\{f_n\colon\, t\in\mathbb N\},$$ where $$f_n(x):=\frac1{1+(x-n)^2}$$ for real $x$. Then $\Phi$ is uniformly bounded in $L^1$, does not escape to vertical infinity, and does not escape to width infinity; however, $\Phi$ is neither tight nor equi-integrable.

Also, $f_n\to0$ pointwise and hence almost everywhere, but $f_n\not\to0$ in $L^1(\mu)$.

$\newcommand\R{\mathbb R}$Your claim is incorrect.

E.g., let $\mu$ be the Lebesgue measure over $X:=\R$. Let $$\Phi:=\{f_n\colon\, t\in\mathbb N\},$$ where $$f_n(x):=\frac1{1+(x-n)^2}$$ for real $x$. Then $\Phi$ is uniformly bounded in $L^1$, does not escape to vertical infinity, and does not escape to width infinity; however, $\Phi$ is neither tight nor equi-integrable.

Also, $f_n\to0$ pointwise and hence almost everywhere, but $f_n\not\to0$ in $L^1(\mu)$.

On the the other hand, it is easy to see that, if a subset $\Phi$ of $L^1(\mu)$ does not escape to vertical infinity and is tight, then $\Phi$ is equi-integrable and uniformly bounded in $L^1$. Also, if $\Phi=\{f_n\colon\, t\in\mathbb N\}$ does not escape to vertical infinity and is tight, and if $f_n\to f$ almost everywhere, then $f_n\to f$ in $L^1(\mu)$.