Let $I$ be a bounded interval in $\mathbb{R}$. Suppose that $f$ is a bounded and Lebesgue measurable function from $I$ to $\mathbb{R}$. Show that the graph has measure 0 with respect to Lebesgue measure in $\mathbb{R}^2$.

I have this (partial and undetailed solution): Let $f(I) \subset J$, where $J$ is a bounded interval. Split J into intervals $J_1,J_2,...J_n$ of lengeth $\leq \frac{1}{n}$. Observe that $f^{-1}(J_k)$ x $J_k$ make up a cover of the graph.

Could someone please help fill in the gaps and turn this into a more detailed solution? I especially don't see why $f^{-1}(J_k)$ x $J_k$is a cover of the graph.

EDIT: For this part, I need to use that I is a bounded interval, I can't just use any measurable function as asked for below and as the answer in the suggested duplicate does. These are two different homework questions and therefore I'm not just supposed to do the second one and for the first one write see below. Also I aleready have this given "solution." I am only interested in an answer of this form; one that fills in the gaps left in the solution I wrote above.

I also have to prove the same for a not necessarily bounded measurable function $f:E \rightarrow\mathbb{R}$, where E is a (not necessarily bounded) measurable subset of R. I am given the hint to exhaust the graph of f by bounded pieces, but I don't quite see what the peices would be


Denote the $\sigma$-algebras of all Lebesgue measureable subsets of $I$ and $\mathbb{R}$ by $\mathcal{L}(I)$ and $\mathcal{L}(\mathbb{R}$) respectively. Dentoe the $\sigma$-algebra of all Borel subsets of $\mathbb{R}$ by $\mathcal{B}(\mathbb{R})$. Let $A=\{(x,y)\mid x\in I\mbox{ and }y=f(x)\}$ be the graph of the functon $f$.

Firstly, we show that $A\in\mathcal{L}(I)\otimes\mathcal{L}(\mathbb{R})$, where $\mathcal{L}(I)\otimes\mathcal{L}(\mathbb{R})$ denotes the product $\sigma$-algebra of $(I,\mathcal{L}(I))$, $(\mathbb{R},\mathcal{L}(\mathbb{R}))$. Let $\pi_{1}:I\times\mathbb{R}\rightarrow I$ and $\pi_{2}:I\times\mathbb{R}\rightarrow\mathbb{R}$ be the canonical projections, defined by $\pi_{1}(x,y)=x$ and $\pi_{2}(x,y)=y$. Define $\theta:I\times\mathbb{R}\rightarrow\mathbb{R}$ by $\theta(x,y)=y-f(x)$. Let $i_{\mathbb{R}}:\mathbb{R}\rightarrow\mathbb{R}$ be the identity map $i_{\mathbb{R}}(y)=y$. Observe that $\theta=i_{\mathbb{R}}\circ\pi_{2}-f\circ\pi_{1}$. Since $\pi_{2}$ is $\mathcal{L}(I)\otimes\mathcal{L}(\mathbb{R})/\mathcal{L}(\mathbb{R})$-measurable and $i_{\mathbb{R}}$ is $\mathcal{L}(\mathbb{R})/\mathcal{B}(\mathbb{R})$-measurable, the composition $i_{\mathbb{R}}\circ\pi_{2}$ is $\mathcal{L}(I)\otimes\mathcal{L}(\mathbb{R})/\mathcal{B}(\mathbb{R})$-measurable. Similarly, $\pi_{1}$ is $\mathcal{L}(I)\otimes\mathcal{L}(\mathbb{R})/\mathcal{L}(I)$-measurable and $f$ is $\mathcal{L}(I)/\mathcal{B}(\mathbb{R})$-mesurable, so the composition $f\circ\pi_{1}$ is $\mathcal{L}(I)\otimes\mathcal{L}(\mathbb{R})/\mathcal{B}(\mathbb{R})$-measurable. It follows that $\theta$ is $\mathcal{L}(I)\otimes\mathcal{L}(\mathbb{R})/\mathcal{B}(\mathbb{R})$-measurable. Finally, $A=\theta^{-1}(\{0\})\in\mathcal{L}(I)\otimes\mathcal{L}(\mathbb{R}).$

Denote the Lebesgue measures on $I$ and $\mathbb{R}$ by $\mu$ and $\nu$ respectively. Since $\mu$ and $\nu$ are both $\sigma$-finite, Fubini-Tonelli Theorem is applicable. By Fubini-Tonelli Theorem, \begin{eqnarray*} \mu\times\nu(A) & = & \int1_{A}(x,y)d(\mu\times\nu)(x,y)\\ & = & \int\left(\int1_{A}(x,y)d\nu(y)\right)d\mu(x). \end{eqnarray*} Note that for each $x\in I$, $1_{A}(x,y)=1$ iff $y=f(x)$. Therefore $\int1_{A}(x,y)d\nu(y)=\nu(\{f(x)\})=0$. Hence, the iterated integral is zero and therefore $\mu\times\nu(A)=0.$

  • $\begingroup$ Actually, $I$ and $f$ need not be bounded. Moreover, $I$ needs not be an interval. The proof works equally well even if $I$ is just a Lebesgue measurable set. $\endgroup$ – Danny Pak-Keung Chan Apr 10 '18 at 0:19

The following is another solution which avoids using the Fubini-Tonelli Theorem. Let $I\subseteq \mathbb{R}$ be a Lebesgue measurable set with $\mu(I)<\infty$, where $\mu$ is the usual Lebesgue measure. Note that $I$ needs not be an interval. Let $f:I\rightarrow\mathbb{R}$ be a bounded Lebesgue measurable function. Denote the graph of $f$ by $A$ (i.e., $A=\{(x,y)\mid x\in I \mbox{ and } y=f(x)\}$). Denote the product Lebesgue measure on $\mathbb{R}^2$ by $\mu\times\mu$. We have that $\mu\times\mu (A) =0$.

Proof: In the first part of the previous proof, we have already shown that $A$ is measurable with respect to the product $\sigma$-algebra $\mathcal{L}\otimes\mathcal{L}$. It remains to prove that $\mu\times\mu (A) =0$. Choose $M>0$ such that $f(I)\subseteq [-M,M]$. Let $\varepsilon>0$ be arbitrary. Then there exists a finite partition $J_1,\ldots,J_n$ of $[-M,M]$ such that $\mu(J_i)<\frac{\varepsilon}{1+\mu(I)}$ for each $i=1,\ldots,n$. (See Remark 1)

More precisely,

  1. For each $i$, $J_i$ is a Borel set,

  2. $J_i\cap J_j=\emptyset$ whenever $i\neq j$,

  3. $[-M,M]=\cup_{i=1}^n J_i$, and

  4. $\mu(J_i)<\frac{\varepsilon}{1+\mu(I)}$ for each $i$.

We assert that $A\subseteq \cup_{i=1}^n f^{-1}(J_i)\times J_i$. Let $(x,y)\in A$, then $y=f(x)\in[-M,M]$. There exists a unique $i$ such that $y\in J_i$. Note that $x\in f^{-1}(J_i)$ and hence $(x,y)\in f^{-1}(J_i)\times J_i$. Also note that $f^{-1}(J_i)\cap f^{-1}(J_j) = f^{-1}(J_i\cap J_j) = \emptyset$ whenever $i\neq j$. Therefore

\begin{eqnarray*} \mu\times\mu(A) &\leq& \sum_{i=1}^n \mu\times\mu( f^{-1}(J_i)\times J_i) \\ & = & \sum_{i=1}^n \mu(f^{-1}(J_i)) \mu(J_i) \\ & \leq & \frac{\varepsilon}{1+\mu(I)} \sum_{i=1}^n \mu(f^{-1}(J_i)) \\ & = & \frac{\varepsilon}{1+\mu(I)} \mu(\cup_{i=1}^n f^{-1}(J_i)) \\ & = & \frac{\varepsilon}{1+\mu(I)} \mu(I) \\ & < & \varepsilon. \\ \end{eqnarray*}

Since $\varepsilon$ is arbitrary, we have $\mu\times\mu(A)=0$.

Remark 1: The partition $\{J_1,\ldots,J_n\}$ exists. In fact, $J_i$ can be chosen as an interval. For, choose $n\in\mathbb{N}$ be sufficiently large such that $\frac{2M}{n}< \frac{\varepsilon}{1+\mu(I)}$. Let $h=\frac{2M}{n}$. Define $J_1 = [-M, -M+h), J_2=[-M+h, -M+2h), \ldots, J_n = [M-h,M]$.

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We further generalize the result by dropping the assumption that $\mu(I)<\infty$.

Proof: For each $n$, let $I_{n}=I\cap[-n,n]$ and $f_{n}=f\vert_{I_{n}}$. Clearly $f_{n}:I_{n}\rightarrow\mathbb{R}$ is a bounded Lebesgue measurable function defined on a Lebesbue measurable set $I_{n}$ with $\mu(I_{n})<\infty$. Denote the graph of $f_{n}$ by $A_{n}$. By the previous discussion, $\mu\times\mu(A_{n})=0$. Note that $A=\cup_{n}A_{n}$. It follows that $\mu\times\mu(A)=0$.

Finally, we drop the assumption that $f$ is bounded.

Proof: For each $n$, let $I_{n}=f^{-1}([-n,n])$ and let $f_{n}=f\rvert_{I_{n}}$. Note that $f_{n}:I_{n}\rightarrow\mathbb{R}$ is a bounded Lebesgue measurable function. Denote the graph of $f_{n}$ by $A_{n}$. By the previous discussion, $\mu\times\mu(A_{n})=0$. Observe that $A=\cup_{n}A_{n}$ and hence $\mu\times\mu(A)=0$.


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