With the extra assumption it is true, and only continuity on $f:[0,\infty)\rightarrow [0,\infty)$ is needed. Of course, it is sufficient to show that some subsequence of $f(u_k)$ converges. So we can also assume w.l.o.g. that $u_k$ converges a.e. to $u$.

Consider the sequence of non-negative measurable functions on $K$, $v_k:=f\circ u_k$. Because $f$ is continuous, it converges a.e. to $v:=f\circ u$, and by the extra hypothesis, $\int_ K v_k \to\int_K v < \infty $. And, as an immediate consequence of the Fatou's lemma, this also implies $\int_ S v_k \to\int_S v$ for every measurable $S\subset K$ (this is the key point, after all, coming from the equality $\int_K v_k =\int_S v_k +\int_{K\setminus S} v_k $: if one does not lose mass globally, one does not lose mas locally).

Now, as a general fact, on a finite measure space $K$ this situation implies the $L^1$ convergence.

Indeed, let be given a number $\epsilon > 0$. There is a number $\delta > 0 $ such that $\int_ S v < \epsilon$ whenever $S\subset K$ has measure $\big| S\big| < \delta$. By the Severini-Egorov theorem $v_k$ converges almost uniformly, so there is some $S\subset K$ of measure less than $\delta$ such that $v_k$ converges uniformly to $v$ on $K\setminus S$. So we have:

$$ \|v _ k - v\| _ {1,K}=\|v _k-v\| _{1,S}+\|v _k-v\| _{1,K\setminus S}$$ $$\le \int _S v _ k + \int _S v + \big|K\big| \|v _ k-v\| _ {\infty,K\setminus S} \le 2\epsilon + o(1)\\ , \quad \mathrm{as}\quad k\to\infty \\ .$$

Since this is true for any $\epsilon > 0$ we have $\limsup_{k\to\infty} \|v _ k - v\| _ {1,K}=0$ proving the convergence . Actually, the finiteness assumption on $K$ may also be dropped.

Here's another proof: consider the sequence $ w_k : = \sqrt v_k \in L^2(K)$. It is norm-bounded, converges to $w : = \sqrt{v}$ a.e., hence weakly in $L^2$; moreover, $\|w_ k \| _ 2 \to \|w \| _ 2 $ . In a Hilbert space, this implies strong convergence (this is immediately seen just expanding $\|w _ k - w\| _ 2 ^2$). The map $L^2\ni w\mapsto w^2\in L^1$ is continuous, and we conclude $v_k \to v$ in $L^1$.

With the extra assumption it is true, and only continuity on $f:[0,\infty)\rightarrow [0,\infty)$ is needed. Of course, it is sufficient to show that some subsequence of $f(u_k)$ converges. So we can also assume w.l.o.g. that $u_k$ converges a.e. to $u$.

Consider the sequence of non-negative measurable functions on $K$, $v_k:=f\circ u_k$. Because $f$ is continuous, it converges a.e. to $v:=f\circ u$, and by the extra hypothesis, $\int_ K v_k \to\int_K v < \infty $. And, as an immediate consequence of the Fatou's lemma, this also implies $\int_ S v_k \to\int_S v$ for every measurable $S\subset K$ (this is the key point, after all, coming from the equality $\int_K v_k =\int_S v_k +\int_{K\setminus S} v_k $: if one does not lose mass globally, one does not lose mas locally).

Now, as a general fact, on a finite measure space $K$ this situation implies the $L^1$ convergence.

Indeed, let be given a number $\epsilon > 0$. There is a number $\delta > 0 $ such that $\int_ S v < \epsilon$ whenever $S\subset K$ has measure $\big| S\big| < \delta$. By the Severini-Egorov theorem $v_k$ converges almost uniformly, so there is some $S\subset K$ of measure less than $\delta$ such that $v_k$ converges uniformly to $v$ on $K\setminus S$. So we have:

$$ \|v _ k - v\| _ {1,K}=\|v _k-v\| _{1,S}+\|v _k-v\| _{1,K\setminus S}$$ $$\le \int _S v _ k + \int _S v + \big|K\big| \|v _ k-v\| _ {\infty,K\setminus S} \le 2\epsilon + o(1)\, , \quad \mathrm{as}\quad k\to\infty \, .$$

Since this is true for any $\epsilon > 0$ we have $\limsup_{k\to\infty} \|v _ k - v\| _ {1,K}=0$ proving the convergence . Actually, the finiteness assumption on $K$ may also be dropped.

Here's another proof: consider the sequence $ w_k : = \sqrt v_k \in L^2(K)$. It is norm-bounded, converges to $w : = \sqrt{v}$ a.e., hence weakly in $L^2$; moreover, $\|w_ k \| _ 2 \to \|w \| _ 2 $ . In a Hilbert space, this implies strong convergence (this is immediately seen just expanding $\|w _ k - w\| _ 2 ^2$). The map $L^2\ni w\mapsto w^2\in L^1$ is continuous, and we conclude $v_k \to v$ in $L^1$.

With the extra assumption it is true, and only continuity on $f:[0,\infty)\rightarrow [0,\infty)$ is needed. Of course, it is sufficient to show that some subsequence of $f(u_k)$ converges. So we can also assume w.l.o.g. that $u_k$ converges a.e. to $u$.

Consider the sequence of non-negative measurable functions on $K$, $v_k:=f\circ u_k$. Because $f$ is continuous, it converges a.e. to $v:=f\circ u$, and by the extra hypothesis, $\int_ K v_k \to\int_K v < \infty $. And, as an immediate consequence of the Fatou's lemma, this also implies $\int_ S v_k \to\int_S v$ for every measurable $S\subset K$ (this is the key point, after all, coming from the equality $\int_K v_k =\int_S v_k +\int_{K\setminus S} v_k $: if one does not lose mass globally, one does not lose mas locally).

Now, as a general fact, on a finite measure space $K$ this situation implies the $L^1$ convergence.

Indeed, let be given a number $\epsilon > 0$. There is a number $\delta > 0 $ such that $\int_ S v < \epsilon$ whenever $S\subset K$ has measure $\big| S\big| < \delta$. By the Severini-Egorov theorem $v_k$ converges almost uniformly, so there is some $S\subset K$ of measure less than $\delta$ such that $v_k$ converges uniformly to $v$ on $K\setminus S$. So we have:

$$ \|v _ k - v\| _ {1,K}=\|v _k-v\| _{1,S}+\|v _k-v\| _{1,K\setminus S}$$ $$\le \int _S v _ k + \int _S v + \big|K\big| \|v _ k-v\| _ {\infty,K\setminus S} \le 2\epsilon + o(1)\\ , \quad \mathrm{as}\quad k\to\infty \\ .$$

Since this is true for any $\epsilon > 0$ we have $\limsup_{k\to\infty} \|v _ k - v\| _ {1,K}=0$ proving the convergence .

With the extra assumption it is true, and only continuity on $f:[0,\infty)\rightarrow [0,\infty)$ is needed. Of course, it is sufficient to show that some subsequence of $f(u_k)$ converges. So we can also assume w.l.o.g. that $u_k$ converges a.e. to $u$.

Consider the sequence of non-negative measurable functions on $K$, $v_k:=f\circ u_k$. Because $f$ is continuous, it converges a.e. to $v:=f\circ u$, and by the extra hypothesis, $\int_ K v_k \to\int_K v < \infty $. And, as an immediate consequence of the Fatou's lemma, this also implies $\int_ S v_k \to\int_S v$ for every measurable $S\subset K$ (this is the key point, after all, coming from the equality $\int_K v_k =\int_S v_k +\int_{K\setminus S} v_k $: if one does not lose mass globally, one does not lose mas locally).

Now, as a general fact, on a finite measure space $K$ this situation implies the $L^1$ convergence.

Indeed, let be given a number $\epsilon > 0$. There is a number $\delta > 0 $ such that $\int_ S v < \epsilon$ whenever $S\subset K$ has measure $\big| S\big| < \delta$. By the Severini-Egorov theorem $v_k$ converges almost uniformly, so there is some $S\subset K$ of measure less than $\delta$ such that $v_k$ converges uniformly to $v$ on $K\setminus S$. So we have:

$$ \|v _ k - v\| _ {1,K}=\|v _k-v\| _{1,S}+\|v _k-v\| _{1,K\setminus S}$$ $$\le \int _S v _ k + \int _S v + \big|K\big| \|v _ k-v\| _ {\infty,K\setminus S} \le 2\epsilon + o(1)\\ , \quad \mathrm{as}\quad k\to\infty \\ .$$

Since this is true for any $\epsilon > 0$ we have $\limsup_{k\to\infty} \|v _ k - v\| _ {1,K}=0$ proving the convergence . Actually, the finiteness assumption on $K$ may also be dropped.

Here's another proof: consider the sequence $ w_k : = \sqrt v_k \in L^2(K)$. It is norm-bounded, converges to $w : = \sqrt{v}$ a.e., hence weakly in $L^2$; moreover, $\|w_ k \| _ 2 \to \|w \| _ 2 $ . In a Hilbert space, this implies strong convergence (this is immediately seen just expanding $\|w _ k - w\| _ 2 ^2$). The map $L^2\ni w\mapsto w^2\in L^1$ is continuous, and we conclude $v_k \to v$ in $L^1$.