# Information about Real Analysis - Part 49

• Title: Riemann Integral for Step Functions

• Series: Real Analysis

• YouTube-Title: Real Analysis 49 | Riemann Integral for Step Functions

• Bright video: https://youtu.be/VdXLenrjIao

• Dark video: https://youtu.be/Ewwk0pHeiQk

• Quiz: Test your knowledge

• Timestamps
• Subtitle in English

1 00:00:00,620 –> 00:00:02,440 Hello and welcome back to

2 00:00:02,450 –> 00:00:04,000 real analysis.

3 00:00:04,489 –> 00:00:05,679 And first, I want to thank

4 00:00:05,690 –> 00:00:06,739 all the nice people that

5 00:00:06,750 –> 00:00:08,199 support this channel on Steady

6 00:00:08,210 –> 00:00:08,960 or paypal.

7 00:00:09,710 –> 00:00:11,489 Now, in today’s part 49 we

8 00:00:11,500 –> 00:00:13,029 will continue talking about

9 00:00:13,039 –> 00:00:14,369 the Riemann integral.

10 00:00:15,229 –> 00:00:16,920 In particular, we will show

11 00:00:16,930 –> 00:00:18,350 that the integral for step

12 00:00:18,360 –> 00:00:20,079 functions is well defined

13 00:00:20,889 –> 00:00:21,389 for this.

14 00:00:21,399 –> 00:00:23,360 Please recall as step function

15 00:00:23,370 –> 00:00:25,149 phi is defined on a compact

16 00:00:25,159 –> 00:00:26,850 interval and it is piece-

17 00:00:26,860 –> 00:00:27,909 wisely constant.

18 00:00:28,840 –> 00:00:30,250 So a typical graph looks

19 00:00:30,260 –> 00:00:30,979 like this.

20 00:00:30,989 –> 00:00:32,598 So we have constant parts

21 00:00:32,610 –> 00:00:34,060 and some jumps involved.

22 00:00:35,139 –> 00:00:36,299 Therefore, the important

23 00:00:36,310 –> 00:00:37,610 thing for us is that we can

24 00:00:37,619 –> 00:00:39,400 choose a partition for the

25 00:00:39,409 –> 00:00:40,639 interval A B,

26 00:00:41,689 –> 00:00:42,840 the explicit definition of

27 00:00:42,849 –> 00:00:44,139 a partition I gave you in

28 00:00:44,150 –> 00:00:45,549 the last video here.

29 00:00:45,560 –> 00:00:47,439 Please recall it’s a finite

30 00:00:47,450 –> 00:00:48,750 set where the elements are

31 00:00:48,759 –> 00:00:50,500 called X with an index.

32 00:00:51,389 –> 00:00:52,819 Now corresponding to such

33 00:00:52,830 –> 00:00:54,650 a partition, we can also choose

34 00:00:54,659 –> 00:00:56,400 numbers, we call C one C

35 00:00:56,409 –> 00:00:57,439 two and so on.

36 00:00:58,299 –> 00:00:59,490 And these are simply the

37 00:00:59,500 –> 00:01:01,290 important values the function

38 00:01:01,299 –> 00:01:02,400 phi can have.

39 00:01:03,119 –> 00:01:03,610 OK.

40 00:01:03,619 –> 00:01:05,138 Then with this, we can

41 00:01:05,150 –> 00:01:06,739 define the integral for the

42 00:01:06,750 –> 00:01:08,139 step function phi

43 00:01:08,900 –> 00:01:10,139 we just calculate the

44 00:01:10,150 –> 00:01:11,599 orientated area of a

45 00:01:11,610 –> 00:01:13,459 rectangle here and sum them

46 00:01:13,470 –> 00:01:15,389 up and then

47 00:01:15,400 –> 00:01:17,169 this finite sum is just

48 00:01:17,180 –> 00:01:19,169 called the integral of Phi

49 00:01:19,709 –> 00:01:21,209 denote it with this

50 00:01:21,220 –> 00:01:21,730 symbol.

51 00:01:22,650 –> 00:01:24,389 Now, the topic for this video

52 00:01:24,410 –> 00:01:26,069 is this proposition here

53 00:01:26,879 –> 00:01:28,410 and it tells us that this

54 00:01:28,419 –> 00:01:29,849 symbol here is well

55 00:01:29,860 –> 00:01:31,400 defined more

56 00:01:31,410 –> 00:01:32,919 precisely, this means no

57 00:01:32,930 –> 00:01:34,209 matter which partition we

58 00:01:34,220 –> 00:01:36,050 choose with this property,

59 00:01:36,150 –> 00:01:37,809 we still get out the same

60 00:01:37,819 –> 00:01:38,330 number.

61 00:01:39,029 –> 00:01:40,129 You see this because on the

62 00:01:40,139 –> 00:01:41,500 left hand side, we don’t

63 00:01:41,510 –> 00:01:42,910 mention the partition at all.

64 00:01:43,930 –> 00:01:45,180 Therefore, in the proof,

65 00:01:45,190 –> 00:01:46,540 we just have to consider

66 00:01:46,550 –> 00:01:48,059 two arbitrarily chosen

67 00:01:48,069 –> 00:01:48,849 partitions.

68 00:01:49,699 –> 00:01:51,029 So let’s call the first one,

69 00:01:51,040 –> 00:01:52,839 P one and the second

70 00:01:52,849 –> 00:01:53,959 one, P two.

71 00:01:54,709 –> 00:01:56,260 So for P one, let’s call

72 00:01:56,269 –> 00:01:57,930 the partition X zero X one

73 00:01:57,940 –> 00:01:59,529 and so on until we reach

74 00:01:59,540 –> 00:02:01,010 the last point XN.

75 00:02:01,819 –> 00:02:03,440 Of course, for P two, we

76 00:02:03,449 –> 00:02:04,860 need different symbols.

77 00:02:04,910 –> 00:02:06,199 Therefore, I would suggest

78 00:02:06,209 –> 00:02:08,119 we take X to the zero.

79 00:02:09,389 –> 00:02:10,399 Otherwise it should look

80 00:02:10,407 –> 00:02:10,958 the same.

81 00:02:10,967 –> 00:02:12,539 So we have X to the one X

82 00:02:12,548 –> 00:02:13,958 to the two and so on, but

83 00:02:13,968 –> 00:02:15,248 the size could be different.

84 00:02:15,259 –> 00:02:16,718 So let’s call it m.

85 00:02:17,889 –> 00:02:19,039 In other words, the number

86 00:02:19,050 –> 00:02:20,669 of elements in the two partitions

87 00:02:20,679 –> 00:02:21,690 could be different

88 00:02:22,699 –> 00:02:22,990 here.

89 00:02:23,000 –> 00:02:24,839 Please recall the whole thing

90 00:02:24,850 –> 00:02:26,029 is a decomposition of the

91 00:02:26,039 –> 00:02:26,639 X axis.

92 00:02:26,649 –> 00:02:28,639 So for example, P one, you

93 00:02:28,649 –> 00:02:30,000 could visualize like this.

94 00:02:31,119 –> 00:02:32,339 On the other hand, P two,

95 00:02:32,350 –> 00:02:33,740 you could visualize in the

96 00:02:33,750 –> 00:02:35,710 same way it could look completely

97 00:02:35,720 –> 00:02:36,630 differently.

98 00:02:36,639 –> 00:02:38,119 But of course, it could also

99 00:02:38,130 –> 00:02:39,179 look similarly

100 00:02:40,289 –> 00:02:41,589 and maybe the first case

101 00:02:41,600 –> 00:02:43,070 we should consider would

102 00:02:43,080 –> 00:02:44,529 be that they are very

103 00:02:44,539 –> 00:02:45,139 similar.

104 00:02:46,039 –> 00:02:47,539 However, here, please don’t

105 00:02:47,550 –> 00:02:49,380 forget both partitions should

106 00:02:49,389 –> 00:02:50,770 work for our chosen step

107 00:02:50,779 –> 00:02:51,660 function Phi

108 00:02:52,559 –> 00:02:53,690 for the first one, we call

109 00:02:53,699 –> 00:02:55,020 the value CJ.

110 00:02:55,130 –> 00:02:56,979 And for the second one, maybe

111 00:02:56,990 –> 00:02:57,699 DJ.

112 00:02:58,589 –> 00:03:00,000 Of course, in some sense,

113 00:03:00,009 –> 00:03:01,259 the values should be the

114 00:03:01,270 –> 00:03:01,910 same.

115 00:03:01,979 –> 00:03:03,850 But in general, the indices

116 00:03:03,860 –> 00:03:05,089 don’t coincide.

117 00:03:05,889 –> 00:03:07,070 The important thing you should

118 00:03:07,080 –> 00:03:08,309 remember here is that the

119 00:03:08,320 –> 00:03:10,009 step function phi is

120 00:03:10,020 –> 00:03:10,949 always the same.

121 00:03:11,669 –> 00:03:13,240 For example, the craft could

122 00:03:13,250 –> 00:03:14,089 look like this.

123 00:03:15,119 –> 00:03:17,100 And then you see one possibility

124 00:03:17,110 –> 00:03:18,339 for a partition would be

125 00:03:18,350 –> 00:03:20,130 X one and X two here.

126 00:03:21,130 –> 00:03:22,460 However, of course, you can

127 00:03:22,470 –> 00:03:23,740 also think of other

128 00:03:23,750 –> 00:03:25,399 possibilities for a partition.

129 00:03:26,470 –> 00:03:27,559 So maybe here on the right

130 00:03:27,570 –> 00:03:29,529 hand side, you just see more

131 00:03:29,539 –> 00:03:31,190 points on the X axis.

132 00:03:32,020 –> 00:03:33,279 We still have points at the

133 00:03:33,289 –> 00:03:34,419 jump points here.

134 00:03:34,429 –> 00:03:35,750 But otherwise we have more

135 00:03:35,759 –> 00:03:36,199 points.

136 00:03:36,210 –> 00:03:37,830 So it’s a finer partition

137 00:03:37,839 –> 00:03:39,160 than on the left hand side.

138 00:03:40,029 –> 00:03:41,639 But otherwise this property

139 00:03:41,649 –> 00:03:43,089 here that is important is

140 00:03:43,100 –> 00:03:43,970 still fulfilled.

141 00:03:44,860 –> 00:03:45,279 OK.

142 00:03:45,289 –> 00:03:47,080 As promised, this is exactly

143 00:03:47,089 –> 00:03:48,399 the first case we want to

144 00:03:48,410 –> 00:03:48,860 consider.

145 00:03:48,869 –> 00:03:50,320 Now more

146 00:03:50,330 –> 00:03:52,000 precisely it means that partition

147 00:03:52,009 –> 00:03:53,610 two just adds more

148 00:03:53,619 –> 00:03:55,080 points to partition one.

149 00:03:55,880 –> 00:03:57,750 This means as a set P

150 00:03:57,759 –> 00:03:59,539 two is a super set of P

151 00:03:59,550 –> 00:03:59,979 one.

152 00:04:01,020 –> 00:04:02,889 In words, we would say partition

153 00:04:02,899 –> 00:04:04,500 two is finer than

154 00:04:04,509 –> 00:04:05,419 partition one.

155 00:04:06,429 –> 00:04:07,820 What this means in the picture

156 00:04:07,830 –> 00:04:09,630 you can see above because

157 00:04:09,639 –> 00:04:11,570 X one is exactly

158 00:04:11,580 –> 00:04:13,410 the same point as X tilda

159 00:04:13,419 –> 00:04:14,050 three.

160 00:04:14,809 –> 00:04:16,760 Moreover X two is

161 00:04:16,769 –> 00:04:18,230 the same point as X two,

162 00:04:18,238 –> 00:04:19,029 the five.

163 00:04:19,760 –> 00:04:21,119 Now on the left hand side,

164 00:04:21,130 –> 00:04:22,649 there is no point between

165 00:04:22,660 –> 00:04:23,329 the two.

166 00:04:23,339 –> 00:04:24,760 But on the right hand side,

167 00:04:24,769 –> 00:04:26,220 we find an additional one.

168 00:04:27,279 –> 00:04:27,649 OK?

169 00:04:27,660 –> 00:04:28,890 Maybe it’s a good idea to

170 00:04:28,899 –> 00:04:30,679 use exactly this example

171 00:04:30,690 –> 00:04:32,179 to explain the general concept

172 00:04:32,190 –> 00:04:32,480 here.

173 00:04:33,329 –> 00:04:34,489 Hence, we can write this

174 00:04:34,500 –> 00:04:35,480 equalities and

175 00:04:35,489 –> 00:04:37,209 inequalities where we have

176 00:04:37,220 –> 00:04:39,079 X one is equal to X to the

177 00:04:39,089 –> 00:04:40,989 three and X to the five is

178 00:04:41,000 –> 00:04:42,109 equal to X two.

179 00:04:43,079 –> 00:04:44,540 And in the middle, we still

180 00:04:44,549 –> 00:04:46,170 find X to the four.

181 00:04:46,869 –> 00:04:48,410 Now, going back to the picture

182 00:04:48,420 –> 00:04:49,730 on the left hand side, we

183 00:04:49,739 –> 00:04:51,190 have our value, we would

184 00:04:51,200 –> 00:04:52,359 call C two

185 00:04:53,299 –> 00:04:54,720 because this is the value

186 00:04:54,730 –> 00:04:56,359 of the second interval here.

187 00:04:57,239 –> 00:04:58,510 However, on the right hand

188 00:04:58,519 –> 00:04:59,869 side, this would be D

189 00:04:59,880 –> 00:05:01,640 four and D

190 00:05:01,649 –> 00:05:02,549 five.

191 00:05:03,859 –> 00:05:05,170 However, the important thing

192 00:05:05,179 –> 00:05:06,950 is this is still equal to

193 00:05:06,959 –> 00:05:07,519 C two.

194 00:05:08,250 –> 00:05:09,989 And of course, this is important

195 00:05:10,000 –> 00:05:11,220 when we want to calculate

196 00:05:11,230 –> 00:05:12,899 the area of the corresponding

197 00:05:12,910 –> 00:05:13,750 rectangles

198 00:05:14,769 –> 00:05:15,059 there.

199 00:05:15,070 –> 00:05:16,470 Please keep in mind on the

200 00:05:16,480 –> 00:05:18,200 right hand side, we have

201 00:05:18,209 –> 00:05:19,459 two rectangles.

202 00:05:20,220 –> 00:05:21,589 So we have to one area on

203 00:05:21,600 –> 00:05:23,059 the left and the other one

204 00:05:23,070 –> 00:05:23,799 on the right.

205 00:05:24,779 –> 00:05:26,190 And maybe not so surprising.

206 00:05:26,200 –> 00:05:27,179 In the end, this should be

207 00:05:27,190 –> 00:05:28,929 the same area as we find

208 00:05:28,940 –> 00:05:30,160 it on the left hand side.

209 00:05:30,970 –> 00:05:32,220 However, the important thing

210 00:05:32,230 –> 00:05:34,109 for us here is this comes

211 00:05:34,119 –> 00:05:35,510 out of the calculation,

212 00:05:36,399 –> 00:05:37,660 namely on the right hand

213 00:05:37,670 –> 00:05:39,299 side, the first rectangle

214 00:05:39,309 –> 00:05:41,160 is the four times the

215 00:05:41,170 –> 00:05:42,380 length of the interval

216 00:05:43,320 –> 00:05:45,100 which is X to the four

217 00:05:45,109 –> 00:05:46,549 minus X to the

218 00:05:46,559 –> 00:05:47,220 three

219 00:05:48,100 –> 00:05:49,839 plus the second rectangle,

220 00:05:49,850 –> 00:05:51,320 which is the five

221 00:05:51,329 –> 00:05:53,040 times the length of the interval.

222 00:05:53,950 –> 00:05:55,619 And this one here is X to

223 00:05:55,630 –> 00:05:57,410 the five minus X to the

224 00:05:57,420 –> 00:05:57,850 four.

225 00:05:58,790 –> 00:06:00,200 And now I see because the

226 00:06:00,209 –> 00:06:01,660 four and the five are the

227 00:06:01,670 –> 00:06:03,450 same, we can factor it

228 00:06:03,459 –> 00:06:05,440 out, then we

229 00:06:05,450 –> 00:06:06,940 can just call it C two

230 00:06:06,950 –> 00:06:08,500 because it’s the same value.

231 00:06:09,489 –> 00:06:10,880 And now we put all the X

232 00:06:10,910 –> 00:06:12,839 tilda into the parentheses

233 00:06:12,869 –> 00:06:14,440 and then you see X tilda

234 00:06:14,500 –> 00:06:16,059 four will cancel out.

235 00:06:16,899 –> 00:06:18,760 Moreover, we also know

236 00:06:18,769 –> 00:06:20,250 X 25 is

237 00:06:20,260 –> 00:06:21,519 exactly X two.

238 00:06:22,220 –> 00:06:23,730 So we can substitute this

239 00:06:23,739 –> 00:06:24,519 immediately.

240 00:06:25,200 –> 00:06:26,769 And the same we can do for

241 00:06:26,779 –> 00:06:28,619 X to the three which is X

242 00:06:28,630 –> 00:06:29,019 one.

243 00:06:30,029 –> 00:06:31,510 Hence, we find this whole

244 00:06:31,519 –> 00:06:33,149 thing here is X two

245 00:06:33,160 –> 00:06:34,929 times X two minus

246 00:06:34,940 –> 00:06:36,799 X one which

247 00:06:36,809 –> 00:06:38,609 is as you know the area

248 00:06:38,619 –> 00:06:40,109 here on the left hand side.

249 00:06:41,160 –> 00:06:41,619 OK.

250 00:06:41,630 –> 00:06:42,980 Now with this calculation,

251 00:06:42,989 –> 00:06:44,779 you should see this works

252 00:06:44,790 –> 00:06:46,579 exactly the same for all

253 00:06:46,589 –> 00:06:47,540 other points here.

254 00:06:48,600 –> 00:06:49,859 In other words, splitting

255 00:06:49,869 –> 00:06:51,540 up the rectangle into more

256 00:06:51,549 –> 00:06:53,130 rectangles here does not

257 00:06:53,140 –> 00:06:54,220 change the area.

258 00:06:55,350 –> 00:06:57,209 Therefore, also summing up

259 00:06:57,220 –> 00:06:59,109 all rectangles gives us the

260 00:06:59,119 –> 00:07:00,100 same result.

261 00:07:00,820 –> 00:07:02,000 So the left hand side here

262 00:07:02,010 –> 00:07:03,480 is the area we calculate

263 00:07:03,489 –> 00:07:04,609 with the first partition.

264 00:07:04,709 –> 00:07:05,839 And on the right hand side,

265 00:07:05,850 –> 00:07:07,640 you see the area we calculate

266 00:07:07,649 –> 00:07:08,720 with the second partition.

267 00:07:09,600 –> 00:07:10,630 And by the argument from

268 00:07:10,640 –> 00:07:12,260 above the sum is the

269 00:07:12,269 –> 00:07:12,690 same.

270 00:07:13,540 –> 00:07:14,609 And this is what we want

271 00:07:14,619 –> 00:07:16,459 to show that the sum does

272 00:07:16,470 –> 00:07:17,839 not depend on the choice

273 00:07:17,850 –> 00:07:18,600 of the partition.

274 00:07:19,480 –> 00:07:21,089 However, please here keep

275 00:07:21,100 –> 00:07:22,739 in mind this only worked

276 00:07:22,750 –> 00:07:24,089 because one partition was

277 00:07:24,100 –> 00:07:25,279 finer than the other.

278 00:07:26,190 –> 00:07:27,579 Therefore, the second case

279 00:07:27,589 –> 00:07:29,230 should be that we don’t have

280 00:07:29,239 –> 00:07:29,649 this.

281 00:07:30,579 –> 00:07:32,279 So we don’t have this subset

282 00:07:32,290 –> 00:07:33,200 relation there.

283 00:07:34,399 –> 00:07:36,380 So P one is not a subset

284 00:07:36,390 –> 00:07:37,179 of P two

285 00:07:38,220 –> 00:07:40,040 and of course, also not

286 00:07:40,049 –> 00:07:41,459 the other way around because

287 00:07:41,470 –> 00:07:42,869 then we would be in the first

288 00:07:42,880 –> 00:07:44,760 case again, it would just

289 00:07:44,769 –> 00:07:46,700 be renaming the partition.

290 00:07:47,649 –> 00:07:49,510 However, in this case, we

291 00:07:49,519 –> 00:07:51,510 still can use the first case

292 00:07:51,519 –> 00:07:52,859 because we could define a

293 00:07:52,869 –> 00:07:54,329 new partition P

294 00:07:54,339 –> 00:07:54,869 three.

295 00:07:55,769 –> 00:07:57,109 It just needs to include

296 00:07:57,119 –> 00:07:58,589 all the points of both

297 00:07:58,600 –> 00:07:59,390 partitions,

298 00:08:00,380 –> 00:08:01,980 which means we have to union

299 00:08:01,989 –> 00:08:03,799 P one with P two.

300 00:08:04,769 –> 00:08:06,690 Now with this P three is

301 00:08:06,700 –> 00:08:08,589 clearly a superset of P one

302 00:08:08,600 –> 00:08:09,380 and P two.

303 00:08:10,359 –> 00:08:12,059 Hence, here in both

304 00:08:12,070 –> 00:08:13,980 cases, we can use the first

305 00:08:13,989 –> 00:08:14,450 case.

306 00:08:15,470 –> 00:08:16,779 Therefore, we immediately

307 00:08:16,790 –> 00:08:18,160 get two results out.

308 00:08:19,040 –> 00:08:20,899 First we get when we calculate

309 00:08:20,910 –> 00:08:22,399 this sum for P one.

310 00:08:22,480 –> 00:08:24,260 So let’s write Sigma with

311 00:08:24,269 –> 00:08:24,940 P one.

312 00:08:25,920 –> 00:08:27,399 Then you know, I just mean

313 00:08:27,410 –> 00:08:29,019 this sum with the correct

314 00:08:29,029 –> 00:08:29,619 partition.

315 00:08:30,649 –> 00:08:32,000 Then the first case tells

316 00:08:32,010 –> 00:08:33,630 us now this is exactly the

317 00:08:33,640 –> 00:08:35,510 same as calculating the sum

318 00:08:35,520 –> 00:08:37,099 with the partition three.

319 00:08:38,030 –> 00:08:38,530 OK.

320 00:08:38,539 –> 00:08:40,330 And a similar result we get

321 00:08:40,340 –> 00:08:42,210 when we apply this inclusion

322 00:08:42,219 –> 00:08:43,330 for the first case

323 00:08:44,169 –> 00:08:45,698 there we get that the sum

324 00:08:45,739 –> 00:08:47,578 using P two is the

325 00:08:47,588 –> 00:08:49,338 same as the sum using

326 00:08:49,348 –> 00:08:50,028 P three.

327 00:08:51,150 –> 00:08:52,739 However, of course, the sum

328 00:08:52,750 –> 00:08:54,299 with P three is always the

329 00:08:54,309 –> 00:08:54,869 same.

330 00:08:54,989 –> 00:08:56,729 So here we see the sum with

331 00:08:56,739 –> 00:08:58,530 P one is exactly the

332 00:08:58,539 –> 00:09:00,250 same as the sum with P two.

333 00:09:01,070 –> 00:09:02,650 Hence also here, the

334 00:09:02,659 –> 00:09:04,570 result is the sum

335 00:09:04,580 –> 00:09:06,289 does not depend on the choice

336 00:09:06,299 –> 00:09:07,090 of the partition.

337 00:09:07,989 –> 00:09:09,409 And this is exactly what

338 00:09:09,419 –> 00:09:10,570 we wanted to prove.

339 00:09:10,580 –> 00:09:12,450 The integral symbol is well

340 00:09:12,460 –> 00:09:13,010 defined.

341 00:09:13,960 –> 00:09:14,409 OK.

342 00:09:14,419 –> 00:09:16,159 So you see this was a technical

343 00:09:16,169 –> 00:09:17,900 proof but we had to do it.

344 00:09:18,619 –> 00:09:20,039 However, now in the next

345 00:09:20,049 –> 00:09:21,440 video, we can talk about

346 00:09:21,450 –> 00:09:23,020 the important properties

347 00:09:23,030 –> 00:09:24,630 the integral has for step

348 00:09:24,640 –> 00:09:25,239 functions.

349 00:09:26,169 –> 00:09:27,349 Therefore, I hope I see you

350 00:09:27,359 –> 00:09:28,789 there and have a nice day.

351 00:09:28,880 –> 00:09:29,549 Bye.

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