The author in a book uses this result.
If $X_n\to X$ in probability and $(X_n)$ is bounded in $L^2(\Omega)$ (that is, $E(X_n^2)<K$ for all $n$), then the sequence $(X_n)$ is uniformly integrable.
My question is:
If we only assume that $X_n\to X$ in probability and that $(X_n)$ is bounded in $L^1(\Omega)$, do we still have that $(X_n)$ is uniformly integrable?
When trying to prove this, I end up using the $L^2$ boundedness. I used that when using Hölder, I highlight when I use it, is it possible to solve it without using Hölder and proving the statement for $L^1$-boundedness, or is it false then? Or can I finish my proof without using Hölder and the $L^2$-boundedness?
Proof-attempt:
First I prove that for all n, and all $\epsilon$, there exists a $k$, such that $P(|X_n|>k)<\epsilon$. By using subsequences and a.s. convergence it is easy to see that $X$ must also be bounded by the same constant in as the sequence in $L^2(\Omega)$. Since $X$ is integrable there is a $k'$ such that $E[|X|1_{|X|>k'}]<\epsilon/2$, by choosing $k'$ bigger than 1, we have that $P(|X|>k')<\epsilon/2$. By convergence in probability, there is an $N$, so that if $n \ge N$, then $P(|X_n-X|>k')<\epsilon /2$. For all $i<N$, since $X_i$ is integrable, there are $k_i$ such that $P(|X_i|>k_i)<\epsilon$. Choose k = $\max\{2k',k_1,\ldots,k_{N-1}\}$. Then $P(|X_n|>k)<\epsilon, \forall n$.
Now I prove the statement. Given $\epsilon$ Since $X$ is integrable, there is a $\delta$, so that if $P(A)<\delta$, then $E[|X|1_A]<\epsilon /2$. By the above argument, there is $k_1$ so that for all $n$ $P(|X_n|>k_1)<\delta$, and there is a $k_2$ such that for all $n$, $P(|X_n|>k_2)<(\frac{\epsilon}{4K})^2$, let $k=\max\{k_1,k_2\}.$
Then for all $n$ we have that $$E[|X_n|1_{|X_n|>k}]\le E[|X|1_{|X_n|>k}]+E[|X-X_n|1_{|X_n|>k}].$$ The first term on the RHS is less than $\epsilon/2$, by the $\delta$, $k_1$. The second term on the RHS is also less than $\epsilon/2$ by Hölder: $$E[|X-X_n|1_{|X_n|>k}]\le E[|X_n-X|^2]^{0.5}P(|X_n|>k_2)^{0.5}\le(\|X_n\|_2+\|X\|_2)\frac{\epsilon}{4K}\le\epsilon /2.$$
Your proof under the $\mathbb L^2$-boundedness is correct. The first paragraph can be shortened a little bit more, noticing that by Markov's inequality, $$\tag{*} \mathbb P\left(\left|X_n\right|\gt k\right)\leqslant k^2\mathbb E\left[X_n^2\right]\leqslant K/k^2.$$
The second part can also be shortened, using Cauchy-Scharz inequality in order to bound $\mathbb E\left[\left|X_n\right|\mathbf 1\left\{\left|X_n\right|\gt k\right\} \right]$ by $ \sqrt{\mathbb E\left[X_n^2\right]} \cdot \sqrt{\mathbb P\left(\left|X_n\right|\gt k\right)} \overset{\mbox{by }(*) }{\leqslant}K/k $. This actually only needs boundedness in $\mathbb L^2$. The convergence in probability is not needed.
The answer given here by carmichael561 gives a counter-example where the assumption on boundedness in $\mathbb L^2$ is relaxed to boundedness in $\mathbb L^1$ (with convergence in probability to $0$) :