Proof: Given functions $f_{j}(x)$ defined on the interval $[a, b]$ and continuous $(j \in N)$, divide the interval $[a, b]$ into $m$ equal parts, then the length of each subinterval is $\Delta x$. Take the left endpoint $\xi_{i} (i=1,2, \cdots, m)$ of each subinterval, then we have
$$\begin{array}{l} \int_{a}^{b}\left|\prod_{j=1}^{n} f_{j}(x)\right| d x=\lim _{n \rightarrow \infty} \sum_{i=1}^{m}\left(\left|f_{1}\left(\xi_{i}\right) \| f_{2}\left(\xi_{i}\right)\right| \cdots\left|f_{n}\left(\xi_{i}\right)\right|\right) \Delta x \\ \int_{a}^{b} f_{j}\left(x_{i}\right) \mid d x=\lim _{n \rightarrow \infty} \sum_{i=1}^{m}\left(f_{j}^{b}\left(\xi_{i}\right) \mid, \quad j=1,2, \cdots n\right. \end{array}$$

Let $\left|a_{1 i}^{n}\right|=\left|f_{1}^{p}\left(\xi_{i}\right)\right|,\left|a_{2 i}^{n}\right|=\left|f_{2}^{n}\left(\xi_{i}\right)\right|, \cdots,\left|a_{n i}^{n}\right|=\left|f_{n}^{n}\left(\xi_{i}\right)\right|$, then the series $\sum_{i=1}^{\infty}\left|a_{1 i}^{n}\right|$, $\sum_{i=1}^{\infty}\left|a_{2 i}^{n}\right|, \cdots, \sum_{i=1}^{\infty}\left|a_{n i}^{n}\right|$ all converge. By the corollary of Proposition 2, we get:
$$\left(\sum_{i=1}^{\infty}\left|a_{i i} a_{2 i} \cdots a_{k i}\right|\right)_{n} \leq \sum_{i=1}^{\infty}\left|a_{i i}^{n}\right| \sum_{i=1}^{\infty}\left|a_{1 i}^{n}\right| \cdots\left(\sum_{i=1}^{\infty}\left|a_{k i n}^{n}\right|\right)$$

That is
$$\left[\sum_{i=1}^{n}\left|f_{1}\left(\xi_{i}\right) f_{2}\left(\xi_{i}\right) \cdots f_{n}\left(\xi_{i}\right)\right|\right]^{n} \leq \sum_{i=1}^{n}\left|f_{1}^{f}\left(\xi_{i}\right)\right| \cdots \sum_{i=1}^{n}\left|f_{n}^{f}\left(\xi_{i}\right)\right|$$

Therefore, we have
$$\left[\sum_{i=1}^{\infty} f_{1}\left(\xi_{i}\right) f_{2}\left(\xi_{i}\right) \cdots f_{\mathrm{n}}\left(\xi_{i}\right) \Delta x\right]^{n} \leq \sum_{i=1}^{\infty} f_{1}^{\prime}\left(\xi_{i}\right) \Delta x \cdots \sum_{i=1}^{\infty} 1_{n}^{\prime}\left(\xi_{i}\right) \Delta x$$

Thus, we obtain:
$$\left[\int_{a}^{b}\left|\prod_{j=1}^{n} f_{j}(x)\right| d x\right]^{n} \leq \prod_{j=1}^{n}\left[\int_{a}^{b}\left|f_{j}^{B}(x)\right| d x\right]$$

Therefore, the original proposition holds.