I've been struggling on this one for hours so if someone could help me I would really really appreciate it!

Let f be a bounded function on [a,b]. Show that the function defined by m(x) = inf{f(w):w in [a,x)} is continuous from the left on (a,b).

I'm really looking forward to seeing your thoughts on this one! Thank you so much!

Originally Posted by Marissa043
Let f be a bounded function on [a,b]. Show that the function defined by m(x) = inf{f(w):w in [a,x)} is continuous from the left on (a,b).
This is an easy to prove question. But it depends upon set theory.
If the function $f$ is bounded and $(a,y)\subset (a,x)\subset [a,b]$ then $\inf\{f(t):t\in (a,x)\}\le\inf\{f(t):t\in (a,y)\}$.

That is, for bounded sets $G \subseteq H \Rightarrow \inf (H) \leqslant \inf (G)$. Thus as defined the function $m$ is non-increasing.

So if $s \in (x - \delta ,x)$ then $[a,s)\subset [a,x)$ and $m(s)\ge m(x)$.

Can you finish?

Well I was trying to use the definition: f is continuous from the left at x0 if f(x0)=limx->x0- f(x). But I'm not really sure what to do from your last step to get there. Can you lead me in the right direction please?

Originally Posted by Marissa043
Well I was trying to use the definition: f is continuous from the left at x0 if f(x0)=limx->x0- f(x). But I'm not really sure what to do from your last step to get there. Can you lead me in the right direction please?
The key here is that $m$ is a non-increasing function.
We know that $\forall x\in(a,b],~m(x)=\inf\{f(t):t\in[a,x)\}$ exists.

If $\varepsilon > 0$, so $\exists f(t)\text{ such that }m(x) where $t\in[a,x)$.

Let $\delta=x-t>0$ if $y\in(x-\delta,x)$ we see that $t=x-\delta.

We know that $m(y)\le m(t).

Now can you finish?