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 $Y_1,\ldots,Y_l\in(L^0)^d$.

Does there exist a $\sigma(1)\in\{1,\ldots,l\}$ such that $$\mu(A\cap\{\lambda_{\sigma(1)}\neq0\})>0?$$

  • $\begingroup$ What are these $l$ lambda's? $\endgroup$ – Fedor Petrov Jun 17 '16 at 5:10
  • $\begingroup$ Any collection of more than one $X_j$ is linearly dependent according to this definition because (if $k=2$, say) we can take $\lambda_1 =1_A X_2$, $\lambda_2=-1_A X_1$. $\endgroup$ – Christian Remling Jun 18 '16 at 2:09

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