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3.) a) Let D and E be measurable sets and f a function with domain D U E. Show that f is measurable if and only if its restriction to D and E are measurable_b.) Let...

Question

3.) a) Let D and E be measurable sets and f a function with domain D U E. Show that f is measurable if and only if its restriction to D and E are measurable_b.) Let f be a function with measurable domain D_ Show that f ismeasurable if and only if the function g defined by g(x) = f (x) for x € D and g (x) 0 for x € D is measurable_

3.) a) Let D and E be measurable sets and f a function with domain D U E. Show that f is measurable if and only if its restriction to D and E are measurable_ b.) Let f be a function with measurable domain D_ Show that f is measurable if and only if the function g defined by g(x) = f (x) for x € D and g (x) 0 for x € D is measurable_



Answers

Find (a) $(f+g)(x),$ (b) $(f-g)(x)$ , (c) $(f g)(x),$ and $(d)(f / g)(x) .$ What is the domain of $f / g ?$ $$f(x)=\frac{1}{x}, \quad g(x)=\frac{1}{x^{2}}$$

So in this question we are told we have a function F of X with the following properties. And we want to answer the following questions about f of X. So first I want to do part D. Because if I sketch part D. If I make a sketch of this function having all these properties, I'll be able to see what's happening with A B and C. So let's make our sketch part D. And then we'll answer A B and C. They say f of X equals zero, has solutions at X equals eight. X equals -4 and no others. So that means I have X intercepts At x equals eight and X equals negative four. And those are my only X intercepts They say the limit as X approaches infinity of F of X is three. And the limit as X approaches negative infinity of F of X is three. That means I have what is called a horizontal A sento At Y. Equals three. They are saying that as I go to the right forever and ever My wise get closer and closer to three and as I go to the left forever and ever my wise are also getting closer and closer 23. They say that the limit as X approaches three from the left of this function is 10. So if I came way up here, Here's The .3, 10. And we are saying that as X gets closer and closer to three But stays to the left of three, my wise get closer and closer to 10. I also know That as X approaches five from the right, my wise approach negative infinity. As X approaches five from the left, my wise approach positive infinity. That means I have a vertical ascent. Okay, At x equals five. And they are saying That as X approaches five from the right, my wise approach negative infinity, I'm gonna move this graph down just a little bit so I can extend upwards. So here is x equals five. And they are saying that as X gets closer and closer to five from the right, my wise go down towards negative infinity As X gets closer to five from the left, my wise approach positive infinity. And finally, My function is continuous for all X values other than five. And so there are no holes, no breaks, no jumps in my graph. And so I think the only way that I could have all of these characteristics is to have a function that roughly speaking looks like this. So that's part D. Now let's go back and answer A B and C in a We want the limit as X approaches three from the right of this function and we're defending our answer by referring to f of X. So we know f is continuous everywhere besides X equals five. So that means specifically the fx is continuous At x equals three if f is continuous at X equals three. The limit As X approaches three from the left of my function has to equal the limit as X approaches three from the right of my function, They said the limit as X approaches three from the left was 10 and so 10 is going to have to be equal to the limit As X approaches three from the right of my function, there's party. Now I need be I want to find f of three and to defend my answer by referring to fx. Well, again they said the fx Is continuous at x equals three. By definition for F to be continuous At x equals three means That the limit as X approaches three of FX Equals there for three. But the limit as X approaches three of this function again is 10. And so I'm getting 10 equals and for three in C I want the sign of F of zero. Is it positive or negative? Well, looking at my graph here, F of zero is positive F of zero is positive. Why do I know this? Well, I know that there is only a zero To the left of that vertical ascent toe to the left of x equals five at X equals what was that? X equals negative four. And I know also F is continuous On the interval from negative 4 to 5. So what this means is since the limit As X approaches five from the left of my function with positive infinity, My f of zero had to be zero. That's my final answer here. Hopefully this makes sense. Have a great rest of your day.

In this problem, we have been asked to prove that the subsets of finite non empty set X are partially ordered by set inclusion. So what we need to do is prove that this relation of set inclusion is a fractional order. So, first of all we need to prove that it is reflexive. This relation will be reflexive because let us consider any subsidy of X. And we can say that he is a subset of E. It is a subset of itself and thus is related to enhance. This relation is reflexive. Next we need to show that it is anti symmetric. Now, for that let us consider any two subsets of X, A and B. And let us assume that he is related to B and B is related to A. So that means that is a subset of B, and B is a subset of A. Now if is included in B and bs included an A. Then the only possible conclusion is that the two sets A and B are equal. So, since this condition implies that is equal to be. Hence we can say that the relation is anti symmetric. And the third condition that we need to show is that it is transitive. So for that let us consider three subsets of X, A, B and C. And let us suppose that is a subset of B, and B is a subset of. See. Now, if A is included in B and bs included in C, then we can conclude that is included in C. For example, if we consider a Venn diagram, this is the set E. This is included in the set B, and that's it is included in the set C. So from this diagram we can see that the set E is included in this set. See So this is what we get. So if it is related to B and B is related to see, we get the day is related to. See. Hence it is transitive. Thus, set inclusion is reflexive, it is anti symmetric, and it is also transitive, thus it is a partial order, and thus we approve that the subsets of a given finite non implicit X are partially ordered by set inclusion.

Suppose you want to find functions F and G. Such that F. S container was at zero and the value of F zero is zero and F G is not continuous at zero. Since F G of X is the same as F of X time strong backs then to make this not continue us, we want to make function G of X That is not continuous at zero since this will affect the product. So how do we make it this container was we just recall that continuity of a function at a point. Let's say X equals C. This is only possible if it Within the three conditions. So the first condition will be for the function G of C to exist. And then the limit as X approaches, see of this function G fx also exist. And then lastly, for this limit as X approaches C of G of X to equal the value of G etc. So if any one of the following do not hold, then it makes it this continuous. So so if you let F of X equal to X squared. That's just container was at zero. Also the value of F at zero here is zero. Then we can make a G of X. A rational expression in which the exponents in the denominator is larger than the exponents of F. So we can say one over X. Cube. So since it didn't say anything about G, then we can form it like this because when you multiply say F G of X, which is the same as F of X G of X. We will then get x squared times one over X cube, which is equal to one over X, which is not continue was at zero. Therefore our functions are x squared and the other one is one over X cube. Get also used one over X to the fourth. As long as it's exponent in the denominator, Fergie is greater than the exponent of F of X. So that when we multiply them, we will still have an X in the denominator, which will Make It. This container was at zero.

The fading function here is if offenses equal do excess square, and here it is shifted just left two times. So it does funnel faxes equal do the fourth X Plus two that is equal toe explosive, too old school and then compressed, particularly, which is by one by four, so that it's equal toe after sex is equal to one by four time former affects that is equal to 14 explosive to hold square, and next it is reflected reflected along X axis. So history affects will be minus minus. Have to affix. It is equal to minus one before in tow, explosive to cool. And finally we're shifting and doubt. By doing so does G effects will be its three or fix minus two. So that was minus one. My four explosive to pull square minus two. So the function in terms of F functionally is geophysics. Music will do Yes, three or fix minus. So let's see the grass of this equations. The first grass is X squared, and the next one is explosive to nexus. One by fourth time, my next cup was won by food times for no. If it's letters explodes to hold square and next door office Negative off one by four times explosion to whisper in the classroom office negative off it is one by four times explodes to hold school minus toe. So let's see the grass off this equation. First good office in this May 2nd roughest, most worth to let the third grounds justice. And the fourth graf is that it's letting one way have to change. Really? Look, to see the complete with us this change it through minus 7 to 3 and then wife remember, people does minus sight to see the graph now. So the blue line is dfx Next. If one affects next, his F two affects Nexus s three affects. Next is F four or fix, which is our Simon immigration.


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