There is a much more general result of Vallee Pousin from which a negative answer to your question follows.

Let $(X,\mu)$ be a measure space. We say that a family of function $\mathcal{F}\subset L^1(X)$ is equi-integrable if for every $\varepsilon>0$ there is $\delta>0$ such that $$ \sup_{f\in\mathcal{F}} \int_E |f|\, d\mu<\varepsilon \quad \text{whenever } \mu(E)<\delta. $$

Theorem (de la Vallee Poussin). Let $(X,\mu)$ be a measure space with $\mu(X)<\infty$ and let $\mathcal{F}\subset L^1(X)$. Then $\mathcal{F}$ is equi-integrable if and only if there is a Young function $\Phi$, $\lim_{t\to\infty}\Phi(t)/t=\infty$ such that $$ \sup_{f\in\mathcal{F}}\int_X\Phi(|f|)\, d\mu\leq 1. $$

Clearly a family consisting of a single function is equi-integrable (by absolute continuity of the integral) so for each $f\in L^1$ there is $\Phi$ with $\Phi(f)\in L^1$.

Also we can adjust the function $\Phi$ around zero arbitrarily without changing integrability of the function $\Phi(f)$ so you can have the condition $\Phi(t)/t\to 0$ as $t\to 0^+$.

You can find the proof of the de la Vallee Poussin theorem in

C. Dellacherie, P.-A. Meyer, Probabilities and potential. C. Potential theory for discrete and continuous semigroups. Translated from the French by J. Norris. North-Holland Mathematics Studies, 151. North-Holland Publishing Co., Amsterdam, 1988.

M.M. Rao, Z.D, Ren, Theory of Orlicz spaces. Monographs and Textbooks in Pure and Applied Mathematics, 146. Marcel Dekker, Inc., New York, 1991.

There is a much more general result of Vallée-Poussin from which a negative answer to your question follows.

Let $(X,\mu)$ be a measure space. We say that a family of function $\mathcal{F}\subset L^1(X)$ is equi-integrable if for every $\varepsilon>0$ there is $\delta>0$ such that $$ \sup_{f\in\mathcal{F}} \int_E |f|\, d\mu<\varepsilon \quad \text{whenever } \mu(E)<\delta. $$

Theorem (de la Vallee Poussin). Let $(X,\mu)$ be a measure space with $\mu(X)<\infty$ and let $\mathcal{F}\subset L^1(X)$. Then $\mathcal{F}$ is equi-integrable if and only if there is a Young function $\Phi$, $\lim_{t\to\infty}\Phi(t)/t=\infty$ such that $$ \sup_{f\in\mathcal{F}}\int_X\Phi(|f|)\, d\mu\leq 1. $$

Clearly a family consisting of a single function is equi-integrable (by absolute continuity of the integral) so for each $f\in L^1$ there is $\Phi$ with $\Phi(f)\in L^1$.

Also we can adjust the function $\Phi$ around zero arbitrarily without changing integrability of the function $\Phi(f)$ so you can have the condition $\Phi(t)/t\to 0$ as $t\to 0^+$.

You can find the proof of the de la Vallee Poussin theorem in

C. Dellacherie, P.-A. Meyer, Probabilities and potential. C. Potential theory for discrete and continuous semigroups. Translated from the French by J. Norris. North-Holland Mathematics Studies, 151. North-Holland Publishing Co., Amsterdam, 1988.

M.M. Rao, Z.D, Ren, Theory of Orlicz spaces. Monographs and Textbooks in Pure and Applied Mathematics, 146. Marcel Dekker, Inc., New York, 1991.

The answer is in the negative and I am just typing a solution.

There is a much more general result of Vallee Pousin from which a negative answer to your question follows.

Let $(X,\mu)$ be a measure space. We say that a family of function $\mathcal{F}\subset L^1(X)$ is equi-integrable if for every $\varepsilon>0$ there is $\delta>0$ such that $$ \sup_{f\in\mathcal{F}} \int_E |f|\, d\mu<\varepsilon \quad \text{whenever } \mu(E)<\delta. $$

Theorem (de la Vallee Poussin). Let $(X,\mu)$ be a measure space with $\mu(X)<\infty$ and let $\mathcal{F}\subset L^1(X)$. Then $\mathcal{F}$ is equi-integrable if and only if there is a Young function $\Phi$, $\lim_{t\to\infty}\Phi(t)/t=\infty$ such that $$ \sup_{f\in\mathcal{F}}\int_X\Phi(|f|)\, d\mu\leq 1. $$

Clearly a family consisting of a single function is equi-integrable (by absolute continuity of the integral) so for each $f\in L^1$ there is $\Phi$ with $\Phi(f)\in L^1$.

Also we can adjust the function $\Phi$ around zero arbitrarily without changing integrability of the function $\Phi(f)$ so you can have the condition $\Phi(t)/t\to 0$ as $t\to 0^+$.

You can find the proof of the de la Vallee Poussin theorem in

C. Dellacherie, P.-A. Meyer, Probabilities and potential. C. Potential theory for discrete and continuous semigroups. Translated from the French by J. Norris. North-Holland Mathematics Studies, 151. North-Holland Publishing Co., Amsterdam, 1988.

M.M. Rao, Z.D, Ren, Theory of Orlicz spaces. Monographs and Textbooks in Pure and Applied Mathematics, 146. Marcel Dekker, Inc., New York, 1991.