I'm trying to find an answer to something. Here &#x03BB;<!-- λ --> denotes the Lebesgue measure

realburitv4

realburitv4

Answered question

2022-05-20

I'm trying to find an answer to something.
Here λ denotes the Lebesgue measure. Furthermore:
f n ( x ) = 1 [ 0 , 1 ] ( 2 m x j )
Where n = 2 m + j ,   0 j 2 m 1. Prove that lim n | f n | d λ 0 and that f n ( x ) doesn't converge pointwise for any x [ 0 , 1 ].
The second part isn't that difficult and not of too much interest. I am rather interested in the first part. My go-to approach would be to define a new function that does the same for x [ 0 , 1 ] and from there we draw a line to zero. This would then be a function which is continuous with compact support. This means that the Lebesgue-Integral would simply be the Riemann Integral.
Let us call this function ϕ n . The problem I run into, however, is that
| f n | d λ = lim n ϕ n d λ = 0 1 | 2 m x j | = | 2 m 1 j |
And thus we get:
lim n | f n | d λ = lim n | 2 m 1 j | 0
What is going on? I would be very grateful for your help!

Answer & Explanation

Alex Baldwin

Alex Baldwin

Beginner2022-05-21Added 4 answers

So 2 m n < 2 m + 1 , m = ln 2 ( n ) , j = n 2 m ,
f n ( x ) = 1 [ 0 , 1 ] ( 2 ln 2 ( n ) x n + 2 ln 2 ( n ) ) ,
and f n ( x ) = 1 if and only if 0 2 ln 2 ( n ) x n + 2 ln 2 ( n ) 1 which is equivalent to
n 2 ln 2 ( n ) 2 ln 2 ( n ) x n 2 ln 2 ( n ) + 1 2 ln 2 ( n ) .
Thus
R f n ( x ) d x = 1 2 ln 2 ( n ) 0.

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