Let $f$ be a non-negative Riemann integrable function on $[a,b]$. If $f$ equals to zero except on an null set,then$\int_a^b f = 0$

Let $A$ be the null set and $M=\sup\left\{f(x):x\in[a,b]\right\}$.

For any $\epsilon>0$, there exists a sequence of intervals $(I_k)$ such that $A\subset \bigcup_{k=1}^\infty I_k$ and $\sum_{k=1}^\infty|I_k|<\epsilon$

My approach is 2-steps.

Step 1: To prove that there exists an interval $I$ such that $\{\bigcup_{k=N}^\infty I_N\} \cap A\subset I$ for some $N$

Step 2: After Step 1, $f$ on $[a,b]\setminus I$ is either zero or nonzero value but is covered by $I$ for some $K<N$, which means those nonzero values can be covered by finite numbers of intervals. Hence, it is easy to find suitable $\delta$ to make the Riemann sum in this step to be smaller than $0.5\epsilon$.

For step 1,

consider $\frac{\epsilon}{2M}$, there is a sequence of intervals $(I_k)$ such that $A\subset \bigcup_{k=1}^\infty I_k$ and $\sum_{k=1}^\infty|I_k|<\frac{\epsilon}{2M}$.

Define $I_k=[a_k,b_k]$, assume that $I_k$ is in the order that $\sup\{f(x):x\in I_k\}\le\sup\{f(x):x\in I_{k+1}\}$, otherwise we rearrange the sequence of interval.

Since $A$ is bounded from $a$ and$b$, $\sup_{x\in[a,b]}A$ is well defined, there exists $x_N\in A$, and thus $x_N\in I_N$ for some $N$ such that $x_N\gt \sup_{x\in[a,b]}A-\frac{\epsilon}{4M}$

Note that $|I_N|\lt \frac{\epsilon}{2M}$, because the infinite sum is less than$\frac{\epsilon}{2M}$, so $a_N\gt sup_{x\in[a,b]}A-\frac{\epsilon}{2M}$, so there exists an interval $I$ such that $I_N \cap A\subset I$. For $N+1$, since $x_N\le \sup\{f(x):x\in I_N\}\le \sup\{f(x):x\in I_{N+1}\}$, there exists $x\ge x_N \ge \sup_{x\in[a,b]}A-\frac{\epsilon}{2M}$, so again $I_{N+1} \cap A\subset I$. Then, we can do induction following the "N+1"argument and yield $\{\bigcup_{k=N}^\infty I_N\} \cap A\subset I$ for some $N$.

$|I|\lt \frac{\epsilon}{2M}$, so the Riemann sum on this interval is less than $\frac{\epsilon}{2}$, combining with setp 2, the proof is done.

If I am right, the assumption that $f$ is Riemann integrable is not needed, and the null set assumption can be weakened to be those intervals only need to be small than any $\epsilon\gt 0$?

If I am wrong, please teach me how to prove it.

Any help would be appreciated.


There must be a mistake in your proof if you nowhere used the assumption that $f$ is Riemann integrable (RI). (I assume we are working within the framework of the Riemann integral.) Otherwise you could let $f$ be the characteristic function of $\mathbb Q \cap [0,1].$ That $f$ satisfies the other hypotheses, but as is well known, it is not RI, hence its integral over $[0,1]$ is not even defined.

The result you're after has a short proof: Let $Z$ be the zero set of $f.$ Then $Z$ is dense in $[0,1].$ Otherwise there is an interval $I$ of positive length such that $Z\cap I = \emptyset.$ It follows that $I\subset A,$ which contradicts the assumption that $m(A)=0.$ Now let $P$ be a partition of $[0,1].$ Then $Z$ intersects each subinterval induced by $P.$ Thus, using the nonnegativity of $f,$ we have

$$0\le L(P,f)\le 0 \implies L(P,f)=0.$$

We then get $\int_0^1 f = \sup_P L(P,f) = 0$ as desired.

We can drop the assumption that $f\ge 0$ if we know the result

$$\int_0^1f = \lim_{\mu(P)\to 0} \sum_{k=1}^{n}f(c_k)\Delta x_k.$$

Here $\mu(P)$ is the mesh size of $P,$ the number of subintervals $I_k$ is $n,$ and the $c_k\in I_k$ can be chosen arbitrarily. In particular, we can choose all $c_k$ to lie in $Z.$ Such Riemann sums would then all be $0,$ showing $\int_0^1f=0.$

  • $\begingroup$ Why$L(P, f)\le0$? $\endgroup$ – Belive Jan 1 '17 at 23:35
  • $\begingroup$ Because $Z$ intersects each subinterval. So the inf over each subinterval is $\le 0.$ $\endgroup$ – zhw. Jan 1 '17 at 23:38
  • $\begingroup$ If I adopt the result in your last paragraph, it seems that I do not need to use the concept of dense set to finish the proof. For each$I_k$ containing an element from the null set, there exists an element not belongs to the null set, as the sum of the intervals is always less than $min |I_k|$ for $I_k$ containing an element from the null set. Hence I can use the elements not belongs to the null set as a tag for each $I_k $, and thus the riemann sum is zero. $\endgroup$ – Belive Jan 1 '17 at 23:49
  • $\begingroup$ I like the simpler proof based on Riemann sum given in your last paragraph. +1 $\endgroup$ – Paramanand Singh Jan 2 '17 at 8:06

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