Yes. The proof is similar to the Borel-Cantelli theorem of probability theory. It can be viewed as a refinement of the standard statement of Borel-Cantelli.
Claim: Let $(X, \mathcal{F}, \mu)$ be a measure space with measure $\mu:\mathcal{F}\rightarrow [0,\infty]$. For each $n\in\{1, 2, 3, ...\}$ let $f_n:X\rightarrow\mathbb{R}$ be a measurable function. Suppose for all $\epsilon>0$ we have
$$ \sum_{n=1}^{\infty} \mu(\{x \in X: |f_n(x)|> \epsilon\})<\infty $$
Then for all $\delta>0$, there is a set $E$ such that $\mu(E)\leq \delta$ and $f_n(x)$ converges uniformly to $0$ for all $x \in E^c$.
Proof: For positive integers $n,k$ define
$$ q_n(k) = \sum_{i=n}^{\infty} \mu(\{x \in X: |f_i(x)|> 1/k\})$$
Comparing with our assumption, if we define $\epsilon=1/k$ then $q_n(k)$ can be viewed as the tail in the infinite sum. The assumption that the infinite sum is finite then implies that for all positive integers $k$ we have
$$\lim_{n\rightarrow\infty} q_n(k)=0 \quad (*)$$
For positive integers $n, k$ define
$$ A_{n,k} = \cup_{i=n}^{\infty} \{x \in X: |f_i(x)|> 1/k\}$$
Then by the union bound:
$$ \mu(A_{n,k}) \leq \sum_{i=n}^{\infty} \mu(\{x \in X: |f_i(x)|>1/k\}) = q_n(k)$$
Fix $\delta>0$. By Equation (*), we know that for each $k \in \{1, 2, 3, ...\}$ we can choose a positive integer $n_k$ such that
$$ q_{n_k}(k)\leq \frac{\delta}{2^k}$$
Define
$$E = \cup_{k=1}^{\infty} A_{n_k,k}$$
Then by the union bound:
\begin{align}
\mu(E) &\leq \sum_{k=1}^{\infty} \mu(A_{n_k,k})\\
&\leq \sum_{k=1}^{\infty} q_{n_k}(k) \\
&\leq \sum_{k=1}^{\infty} \frac{\delta}{2^k}\\
&=\delta
\end{align}
Further, it holds that $f_n(x)$ converges uniformly to $0$ for all $x \in E^c$. This is because
$$ E^c = \cap_{k=1}^{\infty}A_{n_k,k}^c = \cap_{k=1}^{\infty}\cap_{i=n_k}^{\infty}\{x\in X: |f_i(x)|\leq 1/k\}$$
This means that for any $\epsilon>0$, we can choose $k$ such that $1/k\leq \epsilon$ and we know that if $i\geq n_k$ then $|f_i(x)|\leq \epsilon$ for all $x \in E^c$.
$\Box$
Regarding your question "why is this condition meaningful," you can see from the proof that it implies certain tail measures decay to zero. The condition works even when $\mu(X)=\infty$ (the statement of Egorov's theorem uses a different condition that needs $\mu(X)$ to be finite).
