Proving the Zero Lebesgue Measure of Continuous Real Functions: Guidance Needed

[mathmari]

Hey!

I have the following exercise..

Show that the $2$-dimensional Lebesgue measure of the graph of a continuous real function is zero.

Could you give some hints what I could do?? (Wondering)

 

[Euge]

Hey!

I have the following exercise..

Show that the $2$-dimensional Lebesgue measure of the graph of a continuous real function is zero.

Could you give some hints what I could do?? (Wondering)

Hi mathmari,

Let $G$ be the graph of a continuous function $f : A \to \Bbb R$, where $A\subset \Bbb R$. Show that $G$ is a closed subset of $A \times A$. Since $G$ is closed, it is measurable. So you can use the Fubini theorem to show that $m^2(G) = 0$.

 

[mathmari]

Hi mathmari,

Let $G$ be the graph of a continuous function $f : A \to \Bbb R$, where $A\subset \Bbb R$. Show that $G$ is a closed subset of $A \times A$. Since $G$ is closed, it is measurable. So you can use the Fubini theorem to show that $m^2(G) = 0$.

To show that $G$ is a closed subset of $A\times A$ I did the following:

Let $G=\{(x,f(x))|x\in A\}$ be the graph of the continuous function $f$.

Since $f$ is continuous for each convergent subsequence $x_k\rightarrow x$ we have that $f(x_k) \rightarrow f(x)$.

Therefore, each sequence of the form $(x_n, f(xn))\in G$ converges to $(x,f(x))\in G$.

So, the graph is a closed subset of $A\times A$.

Is this correct?? (Wondering)

 

[Euge]

To show that $G$ is a closed subset of $A\times A$ I did the following:

Let $G=\{(x,f(x))|x\in A\}$ be the graph of the continuous function $f$.

Since $f$ is continuous for each convergent subsequence $x_k\rightarrow x$ we have that $f(x_k) \rightarrow f(x)$.

Therefore, each sequence of the form $(x_n, f(xn))\in G$ converges to $(x,f(x))\in G$.

So, the graph is a closed subset of $A\times A$.

Is this correct?? (Wondering)

No, it's incorrect, but you have some good ideas. To show that $G$ is a closed subset of $A\times A$, start by taking an arbitrary limit point $(x,y)\in A\times A$ and show that $(x,y)\in G$, i.e., $y = f(x)$. Being a limit point of $G$, $(x,y)$ is the limit of a sequence $(x_k, f(x_k))$ in $G$. Thus $x_k\to x$ and $f(x_k)\to y$. Use continuity of $f$ to deduce $f(x) = y$.

 

[mathmari]

No, it's incorrect, but you have some good ideas. To show that $G$ is a closed subset of $A\times A$, start by taking an arbitrary limit point $(x,y)\in A\times A$ and show that $(x,y)\in G$, i.e., $y = f(x)$. Being a limit point of $G$, $(x,y)$ is the limit of a sequence $(x_k, f(x_k))$ in $G$. Thus $x_k\to x$ and $f(x_k)\to y$. Use continuity of $f$ to deduce $f(x) = y$.

I got stuck right now... (Worried)

When we take $(x,y)\in A\times A$ and we show that $(x,y)\in G$, do we not show in that way that $A\times A$ is a subset of $G$ ?? (Wondering)

 

[Euge]

I got stuck right now... (Worried)

When we take $(x,y)\in A\times A$ and we show that $(x,y)\in G$, do we not show in that way that $A\times A$ is a subset of $G$ ?? (Wondering)

No, we didn't pick an arbitrary point in $A \times A$, but an arbitrary limit point of $G$, which lies in $A\times A$.

 

[mathmari]

No, we didn't pick an arbitrary point in $A \times A$, but an arbitrary limit point of $G$, which lies in $A\times A$.

Ahaa.. OK!

Let $G$ be the graph of a continuous function $f : A \to \Bbb R$, where $A\subset \Bbb R$. Show that $G$ is a closed subset of $A \times A$. Since $G$ is closed, it is measurable. So you can use the Fubini theorem to show that $m^2(G) = 0$.

Could you explain me how I could use the Fubini theorem to show that $m^2(G)=0$ ?? (Wondering)

 

[Euge]

Could you explain me how I could use the Fubini theorem to show that $m^2(G)=0$ ?? (Wondering)

The $x$-section of $G$ is $G^x = \{y\, :\, y = f(x)\} = \{f(x)\}$. So $m^1(G^x) = 0$ for all $x$. Hence by Fubini's theorem, $m^2(G) = \int m^1(G^x)\, dx = \int 0\, dx = 0$.

 

[mathmari]

The $x$-section of $G$ is $G^x = \{y\, :\, y = f(x)\} = \{f(x)\}$. So $m^1(G^x) = 0$ for all $x$. Hence by Fubini's theorem, $m^2(G) = \int m^1(G^x)\, dx = \int 0\, dx = 0$.

Since the Lebesgue integral is in the next chapter, is there an other way to show it, besides using Fubini's Theorem?? (Wondering)

 

[Euge]

Since the Lebesgue integral is in the next chapter, is there an other way to show it, besides using Fubini's Theorem?? (Wondering)

Before that, let me ask you this. What was the domain of $f$ supposed to be?

 

[mathmari]

Before that, let me ask you this. What was the domain of $f$ supposed to be?

I don't know... (Wondering) The exercise says nothing about the domain of $f$...

 

[Euge]

I don't know... (Wondering) The exercise says nothing about the domain of $f$...

Let's suppose the domain of $f$ is $\Bbb R$, so that $G = \{(x,y)\in \Bbb R^2\, |\, y = f(x)\}$. For each $n\in \Bbb N$, let $G_n = \{(x,y)\in G \, |\, |x| \le n\}$. Then $$G = \bigcup_{n=1}^\infty G_n$$. Using uniform continuity arguments, show that $m^2(G_n) = 0$ for all $n\in \Bbb N$.