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$E$ is separable Hilbert space with norm $|\cdot|$ defined by hermitian form. $f\in L^1(E)$ with $L^1(E)$ integrable functions with target in $E$. For each unit vector $e\in E$,$f_e=\langle f,e\rangle$. Suppose for all measurable $A$ and all $e$, $\big|\int_Af_e\,d\mu\big|\leq b\mu(A)$. Then $|f|\leq b$ a.e.

It is clear that $|f_e(x)|\leq b$ a.e.

$\textbf{Q:}$ How do I deduce $|f|\leq b$ a.e? Since $E$ is separable Hilbert space, I have $f=\sum_e f_e e$. I need $\big|\int_Af\big|\leq b\mu(A)$ for all measurable sets. WLOG, $e$ is a set of orthonormal basis. Then $\int_Af=\int_A f_ee$ this indicates all coordinates $\big|\int_Af_e\big|\leq b$. Now $\big|\int_Af\big|=\sqrt{\sum_e\big|\int_Af_e\big|^2}$ but I might not be able to bound this by $b$.

Ref. Lang, Real and Function Analysis Chpt VI, Sec 6, Cor 5.20

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  • $\begingroup$ $A$ isn't fixed, it should give you a hint that we need to take a proper set $A$. $\endgroup$ – Jakobian Aug 13 at 16:12
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Let $D \subset E$ be countable dense and not containing $0$. Thus, for every $v \in E$, $\|v\|_E=\sup_{a \in D}\,\frac{\langle v,\,a\rangle}{\|a\|_E}$.

There is thus a full measure set $A_0$ such that for all $d \in D$, $x \in A_0$, $\langle f(x),\,d\rangle \leq b\|d\|_E$.

By taking suprema, you conclude that for all $x \in A_0$, $\|f(x)\|_E \leq b$.

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