I am considering the following problem:
Suppose that $\gamma:[0,1] \to K$ is a smooth parametric curve, where $K \subseteq \mathbb{R}^n$ is a compact set. Suppose that $\{f_n\}$ is a sequence of functions from $K$ to $\mathbb{R}$ such that $f_n \in L^1(K)$, for each $n$ and that $$ \lim_{n \to \infty} \| f_n -f \|_{L^1(K)} =0, $$ for some function $f: K \to \mathbb{R}$ in $L^1(K)$. Can we show that $$ \lim_{n \to \infty} \int_0^1 f_n( \gamma(t)) \, dt = \int_0^1 f( \gamma(t)) \, dt \,\,?$$
It appears to me that convergence in the sense of "volume" integrals is a stronger notion than convergence in the sense of "line" integrals, but I can't show this statement rigorously.
NO, this is not true. $f(\gamma(t))$ is not even well defined since $L^{1}$ is an equivalence class of functions. So some smoothness assumption on $f_n$ ad $f$ is required.
Suppose w e fix some elements $f_n$ and $f$ in the equivalence classes. If $\gamma$ is a line segment, for example, then the range of $\gamma$ has $n$ dimensional Lebesgue measure $0$ (if $n >1$). So if we redefine $f_n$ to be $1$ on this segment and $f$ to be $0$ on it then your hypothesis is still satisfied and the conclusion fails.