If F is Riemann Integrable Then is F Continuous

2. if $f$ is Riemann integrable, then $f^2$ is Riemann integrable

if $f$ is Riemann integrable, then $f^2$ is Riemann integrable

Solution 1

Note that if $f$ is not continuous, then $f$ won't necessarily have a minimum or a maximum so you should talk about supremum and infimum. The relevant part that is missing in the proof is the following:

Let $f,g \colon [a,b] \rightarrow \mathbb{R}$ be bounded functions such that $|f(x) - f(y)| \leq |g(x) - g(y)|$. Then $$ \sup_{x \in [a,b]} f(x) - \inf_{x \in [a,b]} f(x) \leq \sup_{x \in [a,b]} g(x) - \inf_{x \in [a,b]} g(x). $$

To prove it, note that

$$ f(x) - f(y) \leq |g(x) - g(y)| = \max \{g(x),g(y) \} - \min \{g(x),g(y) \} \leq \sup_{x \in [a,b]} g(x) - \inf_{x \in [a,b]} g(x) $$

which implies that

$$ \sup_{x \in [a,b]} f(x) - \inf_{x \in [a,b]} f(x) = \sup \{ f(x) - f(y) \, | \, x, y \in [a,b] \} \leq \sup_{x \in [a,b]} g(x) - \inf_{x \in [a,b]} g(x). $$

Solution 2

If $f$ is bounded then $f^2$ is bounded. This should be clear.

Suppose $|f|$ is bounded by $M>0$. Notice that $|f(x)^2-f(y)^2|=|f(x)+f(y)||f(x)-f(y)|$. Consequently $U(f^2,P)-L(f^2,P) \leq 2M \left ( U(f,P)-L(f,P) \right )$. From here it is straightforward to get Riemann integrability of $f^2$.

The above argument can be dinectly adapted to show that if $f$ is Riemann integrable and $g$ is Lipschitz continuous on the range of $f$, then $g \circ f$ is Riemann integrable. I think a similar argument will show that if $g$ is uniformly continuous on the range of $f$ then $g \circ f$ is Riemann integrable, but in this case the details are somewhat more complicated. Note that it is not sufficient to merely assume $g$ is continuous on the range of $f$; for one thing, in this case $g \circ f$ may fail to be bounded.

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Comments

• This was an exercise in my real analysis text and I was a bit confused by the proof. I found it a couple places online and it seems that in general the proof uses the fact that if f is bounded we have that $$|f(x)|\le B $$ for all x $\in$ (a,b) $\rightarrow$

  $$|f^2(x)-f^2(y)|=|f(x)-f(y)||f(x)+f(y)| \le 2B|f(x)-f(y)| $$

  From this they deduce $$RS^+(f^2,p)-RS^-(f^2,p) = \sum_{k=1}^n(f^2(M_k)-f^2(m_k))(x_k-x_{k-1})\le2B(f(M_k)-f(m_k))(x_k-x_{k-1})=2B(RS^+(f,p)-RS^-(f,p))$$ From this it's easy to finish the proof, but the argument up to this point doesn't make sense to me. How do we know that the max and min of $f^2$ is the same min and max that f has on a given interval? Can someone explain this argument to me more in depth? Thanks.

  • An easier proof (although you may not have gotten to these theorems in the course): $f$ is Riemann integrable $\iff f$ is continuous "almost everywhere." (Almost everywhere means everywhere except possibly on a set of measure 0.) Since $f^2$ is the composition of the continuous function $x \mapsto x^2$ and $f$, it is continuous where $f$ is, and hence $f^2$ is continuous almost everywhere, so$f^2$ is Riemann integrable.

  • Haven't learned measure theory yet. Even if that one is easier, I'd still like to understand this one.

• @avid19 You can't define Riemann integration on a non-compact set. But you can adapt your example by making $f : [0,1] \to (0,1]$, $f(x)=x$ for $x>0$ and $1$ otherwise. Then $g : (0,1] \to (0,\infty),g(x)=1/x$ is well-defined on the range of $f$ and $g \circ f$ is not Riemann integrable. So I think the natural condition is instead that $g$ is uniformly continuous on the range of $f$.

• You are quite right

• When you square the max/min of f, does it remain the max/min of $f^2$

• @1233211 The maximum absolute value of $f$, squared, is the absolute value of $f^2$. Similarly the minimum absolute value of $f$, squared, is the minimum value of $f^2$.

• Good point! it's useful

Recents

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Source: https://9to5science.com/if-f-is-riemann-integrable-then-f-2-is-riemann-integrable

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