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I was wondering if anyone could give me any hints on how i would approach this question. In class we went through a solution that included truncating the function $$f_k = \text{min}\{f,k\}$$ for some bound k and taking some limits with it. However I was wondering if there were any other methods of tackling the problem. Perhaps using Dominated convergence with $f \cdot 1_{C_n} \leq f$ and taking the limit? Or using absolute continuity?

Let $f \in M^{+}(X, X, \mu)$ be a positive measurable function such that the integral is finite: $\int f d \mu<\infty$.

Let $\left(C_n\right) \subset \mathbb{X}$ be a sequence of measurable sets such that $\mu\left(C_n\right) \rightarrow 0$.

I want to show that:

$$ \int_{C_n} f d \mu \rightarrow 0 . $$

veganwithabeef
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1 Answers1

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One can indeed use dominated convergence (version with convergence in measure). We have $g_n:=f\mathbf{1}_{C_n}\leq f \in L^1$ uniformly in $n$. Now let $\varepsilon>0,A\in \mathscr{A}$ with $\mu(A)<\infty$. We get $$ \begin{aligned} \mu(\{x:g_n(x)>\varepsilon\}\cap A)&\leq \mu(\{x:f(x)>\varepsilon\}\cap C_n\cap A)\\ &\leq \mu(C_n)\stackrel{n\to \infty}\to 0 \end{aligned}$$ Thus $g_n\to^\mu 0$ so that by DCT $\int g_nd\mu\to 0$.

Snoop
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