How to prove that there exists $\lambda_{\sigma(1)}$ such that $\mu(A\cap\{\lambda_{\sigma(1)}\neq0\})>0$? Let $(\mathcal F,\Omega,\mu)$ be a measure space and $A\subseteq\Omega$ such that $\mu(A)>0$. Let $L^0$ be the space of all measurable functions. 
We say $X_1,\ldots,X_k\in(L^0)^d=\prod_{k=1}^dL^0$ are linearly independent on $A$ if $(0,\ldots,0)$ is the only vector $(\lambda_1,\ldots,\lambda_k)\in1_A(L^0)^d$ satisfying
$$\lambda_1X_1+\ldots+\lambda_kX_k=0$$
Suppose $X_1,\ldots,X_k\in(L^0)^d$ are linearly independent on $A$ and
$$\text{span}_A\{X_1,\ldots,X_k\}\subseteq\text{span}_A\{Y_1,\ldots,Y_l\}$$
for some $Y_1,\ldots,Y_l\in(L^0)^d$. 
I'm  trying to prove that there exists a $\sigma(1)\in\{1,\ldots,l\}$ such that 
$$\mu(A\cap\{\lambda_{\sigma(1)}\neq0\})>0$$
I tried to conclude this from the fact we can write
$$1_AX_1=\sum_{i=1}^l\lambda_i1_AY_i$$
while couldn't arrive to any result, Can somebody help me, please?

Suppose, this is not true. Then, we would have $\mu(A\cap\{\lambda_i\ne0\})=0$ for all $i\in\{1,2,\dotsc,l\}$. Hence,
$$1_A X_1 = \sum_{i=1}^l \lambda_i1_AY_i$$
implies that $1_AX_1=0$. This, the vector $(1_A,0,\dotsc,0)$ satisfies
$$1_AX_1 + 0X_2 + \dotsb + 0X_k = 0$$
which contradicts the linear independence of $(X_1,X_2,\dotsc,X_k)$.
