I am able to cook up an example A sequence in $L^p$ that converges in measure but not weakly. Namely, let $f_n = n \chi_{[0, 1/n]}$ for $n \in \mathbb{N}.$ Then we have that $\{f_n \}$ converges in measure on $[0, 1]$ but it does not converge weakly in $L^p([0, 1]).$
My question is can we cook up an example for the reverse? i.e. can we find a sequence of functions in $L^p$ that converges weakly but not in measure?
I am under the impression that this question has been studied before here at MSE. I was not able to find a duplicate question however. In any event, here is an example of a sequence that solves the OP. I leave some details to the OP (see Comments).
Let $\phi$ be a $1$-periodic bounded measurable function such that $$\int^1_0\phi(x)\,dx=1,\qquad \int^1_0|\phi(x)|\,dx>0$$ For each $n\in\mathbb{N}$, define $\phi_n(x)=\phi(nx)$. By Fejer's theorem, for any $f\in L_1(\mathbb{R})$, $$\lim_n\int_{\mathbb{R}}f(x)\phi(nx)\,dx =\Big(\int^1_0\phi(x)\,dx\Big)\Big(\int_{\mathbb{R}}f(x)\,dx\Big)=0$$ In particular, as any $f\in L_1(0,1)$ can be extended as a function in $L_1(\mathbb{R})$ by setting $f(x)=0$ outside $(0,1)$, we have that $$\lim_{n\rightarrow\infty}\int^1_0f(x)\phi(nx)\,dx=0$$ This shows that $\{\phi_n:n\in\mathbb{N}\}$ converges weakly to $0$ in $L_1(0,1)$ as $n\rightarrow\infty$. Since $L_q(0,1)\subset L_1(0,1)$ for all $q\geq 1$, and $\phi_n\in L_p(0,1)$ for all $1\leq p\leq \infty$, the sequence $\{\phi_n\}$ converges weakly to $0$ in any $L_p(0,1)$, $1\leq p<\infty$.
Finally, as $$\int^1_0|\phi(nx)|\,dx=\frac1n\int^n_0|\phi(t)|\,dt=\int^1_0|\phi(t)|\,dt>0$$ we have that $\phi_n$ does not converge to $0$ in probability (why?).
Comments:
When $\phi_{\pm}(x)=e^{\pm 2\pi inx}$, Fejer's theorem yields the Riemann-Lebesgue lemma.
A similar result can be obtained by considering the real or imaginary parts of $\phi_{\pm}$, namely $\cos(2\pi x)$ and $\sin(2\pi x)$.
Another interesting example is the $1$-periodic function $\phi$ defined as $\phi(x)=\mathbb{1}_{(0,1/2)}-\mathbb{1}_{(1/2,1)}(x)=\operatorname{sign}\big(\sin(2\pi x)\big)$ for $0\leq x\leq1$.
There is an interesting posting related to the OP which gives conditions convergence in measure implies weak convergence.