# show that $\|f\|_{\infty} = \sup_{x \in [0,1]} |f(x)|$ if $f$ is continuous.

1. Show that if $f :[0,1] \to \mathbb{R}$ is continuous, then $\|f\|_{\infty} = \sup_{x \in [0,1]} |f(x)|$ with respect to lebesgue measure.

2. Show that this can fail for other measures.

I am quite new to measure theory, and I don't know how to begin these exercises.

My textbook defines $\|f\|_\infty$ as the essential supremum of $f$.

So, I think that $|f(x)| \le \|f\|_{\infty}$ a.e. on $[0,1]$, and "a.e." implies that we remove the set, for which $f$ is discontinuous at elements in this set. But, how can I show the equality?, and I don't understand how different types of measures affect the answer.

Could you give some help?

You want to show that the essential supremum of $f$ is equal to the supremum of $f$ on the interval $[0,1]$.

While this might seem obvious, it is not because the essential supremum is a bound on $f$ "outside" a set of measure zero, so it is possible that if $f$ behaves badly on this set of measure zero, then the supremum of $f$ could be very high.

The definition of essential supremum is that it is the smallest "essential upper bound" i.e. the infimum of all essential upper bounds, which are numbers that are a.e. bounds of $|f|$ on $[0,1]$. Mathematically, let $U = \{a \in \mathbb R : |f| \leq a \ \mbox{a.e.}(\mu)\}$, then $U$ has a lower bound, so has a greatest lower bound(infimum), and $||f||_{\infty} = \inf U$.

How to show that $||f||_{\infty} = \sup_{[0,1]} |f|$?

We must obviously use continuity of $f$. Let $L = ||f||_{\infty}$ and $M = \sup_{[0,1]} |f|$.

By continuity, we know that there exists $y \in [0,1]$ such that $f(y) = M$ by the extreme value theorem. Note that $M \in U$ (defined earlier), since $|f| \leq M$ everywhere (therefore almost everywhere), so $L \leq M$ is clear.

The other part, let $\epsilon > 0$. We will show that $M-\epsilon \notin U$. Consequently, by definition of infimum, we get $M = \inf U = L$.

By continuity at $y$ (mentioned earlier) we get $\delta > 0$ such that $|x-y| < \delta \implies |f(x) - M | < \epsilon \implies |f(x)| > M - \epsilon$.

Therefore, we see that $(y-\delta,y+\delta)\subset \{|f| > a\}$. Here we use monotonicity of the Lebesgue measure to conclude that $\mu(|f| > a) \geq \mu((y-\delta,y+\delta))$ and then the definition on intervals to conclude that $\mu(|f| > a) \geq 2\delta > 0$, so that $M - \epsilon \notin U$. (I have highlighted which properties of the Lebesgue measure we have used. Note that any other measure must not have both these properties).

To construct a counterexample, take the favourite point measure : $\mu(A) = 0$ if $0 \notin A$ and $\mu(A) = 1$ if $0 \in A$. (Check this is a measure : very easy to see this). Call this measure $\delta$, which makes more sense than at first sight (it is the Dirac delta measure).

Under this, the function given by $f(x) = x$ has essential supremum zero. Why? Because for example, $|f| < 0.1$ almost everywhere $\delta$, since the set on which $|f| \geq 0.1$ does not include zero, so has measure zero. Nothing special about $0.1$: you can take anything close to zero to see that $0$ is the essential supremum of $f$. On the other hand, the supremum of $f$ is $1$, so it is clear that equality does not hold. Note that $||f||_{\infty} \leq\sup |f|$ will always hold, because the empty set must always have measure zero.

Note that the Dirac measure does not satisfy the "intervals" property of the Lebesgue measure : the measure of $[0,\frac 12]$ is not the same as the measure of $[\frac 12,1]$. But it is monotone.

• Thanks you so much. I want to give more than one upvote. – Daeseon Sep 14 '18 at 7:26
• You are welcome! – астон вілла олоф мэллбэрг Sep 14 '18 at 8:08
• For what it's worth: $\|f\|$ ($\|f\|$) tends to look better than $||f||$ ($||f||$), at least in my opinion – Omnomnomnom Sep 14 '18 at 12:44
• I will keep this in mind the next time I use it. Thank you @Omnomnomnom – астон вілла олоф мэллбэрг Sep 14 '18 at 13:14