# “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

• A sort of counterexample occurs when one assumes only finite additivity rather than countable additivity. Suppose $X=\{1,2,3,\dots\}$ and the measure of a subset is the "density" of the set. Then the integral of the everywhere positive function $x\mapsto 1/x$ is $0$. – Michael Hardy Sep 21 '11 at 18:08

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.

• Gortaur: Thanks for the answer. I have written that $\int_X v d\mu$ exists in the sense that it is not equal to $\infty$. The proof is neat. From this we obtain that $\int_X (v-u) d\mu >0$. Can we say from this that $\int_X v d\mu> \int_X u d\mu$? – jpv Sep 21 '11 at 14:21
• @jpv: You're welcome. Note, that $\int (v-u)d\mu = \int vd\mu-\int ud\mu$ so $\int (v-u)d\mu>0$ iff $\int vd\mu - \int ud\mu >0$, so $\int vd\mu > \int ud\mu$. I would also add then when you're dealing with strict inequalities, this trick with $X_n$ is usually useful. – Ilya Sep 21 '11 at 14:25
• Gortuar: Thanks, I understood the logic involved here. – jpv Sep 21 '11 at 14:30

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.

• Didier: Thanks for the answer. Both the answers followed very similar steps. Seems that I can only select one correct answer so I cannot tick this even though it is correct! – jpv Sep 21 '11 at 14:31