# "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 $0$. Commented Sep 21, 2011 at 18:08
• Anyone know of a direct proof for the equivalent statement for the Riemann integral? (An indirect proof is the following: when the Riemann integral exists, it coincides with the Lebesgue integral, and so we can use the proofs provided). The proofs below don't work for Riemann integrals for several reasons (i.e., we don't know if those intermediate domains/functions are Riemann-integrable). Commented Dec 18, 2023 at 18:49

Edit (after 11 years): Somehow nowhere (in the OP or in the previous version of the answer) I see being mentioned that strict integral inequality only holds if $$v>u$$ on some set of strictly positive measure.

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
Commented Sep 21, 2011 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.
– SBF
Commented Sep 21, 2011 at 14:25
• Gortuar: Thanks, I understood the logic involved here.
– jpv
Commented Sep 21, 2011 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
Commented Sep 21, 2011 at 14:31
• I believe you meant "by contradiction" en.wikipedia.org/wiki/Proof_by_contrapositive Commented Sep 14, 2022 at 8:27