TITLE: A Quick Question on the Monotone Property of Integrals.
QUESTION [0 upvotes]: Let $(\Omega,\mathcal{A},\mu)$ be a measure space with $f$, $g\in \mathcal{L}_1(\Omega,\mathcal{A},\mu)$.  If for any $A\in\mathcal{A}$ we have $$\int_Afd\mu\geq\int_Agd\mu\space ,$$
show that $f\geq g\space a.e.$  I guessing this must be quiet straightforward, but I am a bit stuck and can't see how I can show $\mu(f<g)=0$ any help and assistance is greatly needed.  Thanks in advance

REPLY [0 votes]: Suppose that it is not true that $ f \geq g $ almost everywhere. Then by definition,
\begin{align}
S & \stackrel{\text{df}}{=}
    \Omega \setminus \{ x \in \Omega \mid f(x) \geq g(x) \} \\
& = \{ x \in \Omega \mid f(x) < g(x) \} \\
& = \{ x \in \Omega \mid g(x) - f(x) > 0 \}
\end{align}
is a measurable subset of $ \Omega $ with non-zero measure. Define a partition $ (S_{n})_{n \in \mathbb{N}} $ of $ S $ by
$$
\forall n \in \mathbb{N}: \quad S_{n} \stackrel{\text{df}}{=}
\left\{ x \in S ~ \middle| ~ g(x) - f(x) > \frac{1}{n} \right\}.
$$
Then $ \mu(S_{n}) > 0 $ for some $ n \in \mathbb{N} $, which implies that
$$
\int_{S_{n}} (g - f) ~ \mathrm{d}{\mu} \in (0,\infty].
$$
However, we know that $ f,g \in {\mathcal{L}^{1}}(\Omega,\mathcal{A},\mu) $, so
$$
\int_{S_{n}} g  ~ \mathrm{d}{\mu} > \int_{S_{n}} f ~ \mathrm{d}{\mu},
$$
which contradicts your assumption for the choice $ A = S_{n} $.
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