"Strict" Monotonicity of Integral

Let $(X,\mathcal{E},\mu)$ be a measure space. Let $u,v$ be $\mu$-measurable functions. If $0 \leq u \leq v$ and $\int_X v d\mu$ exists we know that $\int_X u d\mu \leq \int_X v d\mu$.

I wanted to know if $0 \leq u < v$ and $\int_X v d\mu$ exists then is it true that $\int_X u d\mu < \int_X v d\mu$? This can be shown for simple functions easily i.e. if $u,v$ are simple.

I have assumed here that $\int_X u d\mu = \sup \{\int_X u_n d\mu, u_n \text{simple}, \mu-\text{measurable}, u_n \leq u\}$ where a simple function is defined to be a function whose cardinality of the range is finite.

Any help is greatly appreciated.

Thanks, Phanindra

Solution 1:

First of all, $\int_X v\,d\mu$ exists since $v\geq 0$. Second, you want to show that $\int\limits_X(v-u)\,d\mu>0$ where $v-u$ is an arbitrary positive measurable function, so it's the same as to ask if $\int\limits_Xf\,d\mu>0$ for a positive function $f$ (of course, measurable).

Consider the set $X_n = \{x\in X:f(x)\geq\frac1n\}$ for all $n\in \mathbb N$. Then $\bigcup\limits_{n=1}^\infty X_n = X$ and $X_n\subseteq X_{n+1}$.

Now, suppose that $\mu(X)>0$ and $$ \int\limits_Xf\,d\mu = 0 $$ so $$ 0 = \int\limits_Xf\,d\mu \geq\int\limits_{X_n}f\,d\mu\geq\frac1n\mu(X_n) $$ so $\mu(X_n) = 0$ for any $n\in\mathbb N$. By the continuity of measure we obtain that $$ \mu(X) = \mu\left(\bigcup\limits_{n=1}^\infty X_n\right) = \lim\limits_{n\to\infty}\mu(X_n) = 0 $$ which is not true.

Solution 2:

By contraposition, you might want to prove that if $w\ge0$ and $\displaystyle\int\limits_Xw\mathrm d\mu=0$ then $w=0$ $\mu$-almost everywhere. To see this, consider $A_n=\{x\mid w(x)\ge1/n\}$ and note that $w\ge n^{-1}\mathbf 1_{A_n}$ hence $\displaystyle\int\limits_Xw\mathrm d\mu\ge n^{-1}\mu(A_n)$ hence $\mu(A_n)=0$ for every $n$ hence $\{x\mid w(x)\ne0\}=\bigcup\limits_nA_n$ has measure zero. You are done.