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Let $(X, \mathcal{B}, \mu)$ be a measure space alongside measurable $f:X \rightarrow \mathbb{\bar{R}}$ and suppose $T \in \mathcal{B}$ s.t. $\mu(T) > 0$ and that $\forall t \in T$, $f(t) > 0$. Then I want to show that there exists a simple function $\phi:X \rightarrow \mathbb{\bar{R}}$ s.t. $0 \le \phi \le f$ whereby $\int_T \phi > 0$. For simplicity assume furthermore that $\forall x \in X$, $\phi(x) = 0$ if $x \notin T$.

Working by way of contradiction, it follows that if such a $\phi$ did not exist, $\forall \phi$ s.t. $0 \le \phi \le f$ we would have that $\phi = 0$ almost everywhere. My idea is to then show that it would have to be that given this $f$ itself would have to be equal to 0 almost everywhere (a contradiction). I can't think of how to do this, but feel that this is probably a too circuitous and long-winded way of showing something that could be done much simpler anyway. Any advice?

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Let $A_n = \{ f \ge 1/n\}$. Observe that $\bigcup_n A_n = T$ and conclude that $\mu(A_n) > 0$ for some $n$. Take $\phi = \frac{1}{n} 1_{A_n}$.